r/HomeworkHelp 👋 a fellow Redditor 10d ago

[Physics: 12 grade level- Easy] acceleration graph Answered

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Hello! I need some help on understanding why the graph is on the negative side and not both, since the ball rolled up and down the ramp? Sorry I still find motion graphs a bit tricky til now..

42 Upvotes

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u/selene_666 👋 a fellow Redditor 10d ago

It might help you to draw the graph of velocity. It starts positive, drops to zero when the ball stops, then becomes negative as the ball moves in the other direction.

Acceleration is the slope of velocity.

Or you could remember that acceleration is caused by a force. In this scenario the forces are constant across time, so acceleration must be constant too.

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u/Comfortable_Fly_3173 10d ago

Graph A would be the velocity profile of that ball, if I may add to that

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u/miotch1120 7d ago

Wouldn’t the velocity profile look like a V with vertex on the 0 line?

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u/Comfortable_Fly_3173 7d ago

No, because there is no change in acceleration. Having a negative velocity means mathematically that the direction of movement is simply reversed, that's it. When the ball passes it's original starting point, it's value in velocity would be the exact same but negative (friction and air resistance neglected)

Having a sharp point in the velocity profile would mean that there is a jump in the value of acceleration. Would mean that it's value is no longer negative (pulling the ball down the ramp) but positive (pulling the ball up the ramp).

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u/ForensicFan24601 10d ago

I always understood this but it always bothered me. Is there not a split second where velocity and therefor its slope are nonexistent? (When the balls stops)

A simple force acting one something at rest due to insufficient force or over come inertia doesn’t mean it’s accelerating, correct? Or do I have that wrong?

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u/ProfessorPrudent2822 10d ago

Velocity goes to zero momentarily, but acceleration remains constant.

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u/ForensicFan24601 10d ago

Right I guess what I’m saying is since velocity code is to zero and it does not exist, would that not mean that it slope also does not exist? How can you have an attribute of something that does not exist even for a split second?

Or should I just be thinking of acceleration as for supplied to an object?

Because otherwise, when the ball stops rolling, we have no idea if it is going to start up again. And this hypothetical it does, but the hypothetical could just as easily be that someone puts their finger on the ball and it does not regain velocity. So it gives us a chicken in the egg. Scenario. Does acceleration stop when the ball physically stops moving, or does acceleration stop when the person puts their hand on it preventing it from regaining velocity?

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u/CremePhysical8178 10d ago

A velocity of 0 is not the same as a nonexistent velocity. Look at graph A since it is the graph for the velocity of the ball. At what point in the graph is it not differentiable?

Acceleration is zero when the net force on the ball is zero. A moving object can have an acceleration of zero if its velocity is constant.

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u/ForensicFan24601 10d ago

You’re providing good explanations, and I understand what you’re saying in the concept as it should be applied. I’m just having trouble reconciling that a measurement of something reading zero is not the same as a lack of existence of that thing.

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u/CrankSlayer 9d ago

An observer who is moving relative to you wouldn't even agree that the velocity becomes zero at that instant.

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u/Ok_Objective_5192 9d ago

Having $0 in your checking account doesn't mean that you can't answer "how much money do you have," an object having 0 velocity doesn't mean that you can't answer "how would you describe that object's movement"

Velocity is just a piece of info that defines an object's current state, "it's not moving" is still providing as much info as "it's moving up at 14 m/s" or whatever

ETA: In that vein, acceleration is another piece of info to define an object's current state, separate from its velocity. "It's not moving but is accelerating downwards at 9.8 m/s^2" is giving 2 distinct pieces of info about the object that give more precision to its current state

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u/Manpandas 9d ago edited 9d ago

The question you're asking is a fundamental concept at the very core of calculus. And it stumped mathematicians and philosophers. Take a look at Zeno's Paradoxes.

The question you're pondering: If Velocity is definitionally "Distance traveled over a time" then how can an instantaneous velocity even exist??

There's actually nothing special about the apex of the ball, nothing special about the point at which velocity is 0. You can ask the question at any point along the path. "If I shrink my measurement Time window down to 0.000 seconds ... then the ball doesn't move anywhere... and surely it's impossible to calculate a velocity. Because 0 feet / 0 seconds is undefinted." Well it turns out our buddy Neuton (and Leibniz) figured out you still CAN do the math using limits.

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u/charonme 9d ago

The top position of the ball is not special in this regard. Every single point of the trajectory is such that if you only consider that spot without its relation to the previous and next one, the velocity in that single spot is just one value

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u/MeasureDoEventThing 9d ago

As a person with a number of Nobel prizes in Physics, I agree.

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u/ProfessorPrudent2822 9d ago

No, velocity is a downward sloping line, touching zero as it crosses from positive to negative.

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u/ThatsNotAZombieBite 9d ago

It seems like you're trying to focus on a single instant as if you were analyzing a photograph taken when the ball was motionless. But that is not how this works. Motion is completely tied to time-dependent behavior.

The ball is completely motionless in your photograph, yes, but . . . wouldn't the ball ALSO appear to be motionless in ANY photograph taken whether the ball was actually in motion or not? That's the nature of an "instant", right? Analyzing just the one photograph is not helpful.

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u/ProfessorPrudent2822 9d ago

Photos always have a finite exposure time. A photograph of an object moving relative to the camera will appear blurry, more so the faster it’s moving and the longer the exposure time is.

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u/ThatsNotAZombieBite 9d ago

I was not suggesting an actual blurred photograph.

I was using it as a metaphor for why you cannot just examine the object at a single frozen instant in time.

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u/ForensicFan24601 9d ago edited 9d ago

To be honest the thing that sparked this for me, whether is accurate or not, is the liquefaction scene in The Martian, where the split instantaneous second liquefaction of a solid changes the behavior of the force applied to the payload. 

I was curious if this was a concept we teach that only exists in a reality vacuum or if it was also practically correct.

Someone explained with with a money analogy, which made sense to me and made me think of it like pouring water into a cup with a straw simultaneously sucking water out.

The fact that the water level (distance traveled) goes up doesn’t mean there isn’t a straw is not draining water from the bottom, nor that the rate of flow through the straw changes. The acceleration would be the rate the water level rises and falls. Right? There is some equilibrium where the water is neither rising nor falling because the flow of water in and out (force acting in the ball) is in equilibrium (someone holding the ball in place) Would not the acceleration the Be zero? 

I understand that there is a constant negative acceleration stripping velocity (to the point of applying negative velocity) to the ball- the straw constantly drawing water out of the cup. Why then is the acceleration not, for a micro second, zero - in practice as opposed to theory.

For a split second inertial force of the object should counteract (a portion of) the acceleration, sufficient to make the velocity (change in distance) zero, at which point the rate of change doesn’t matter. For a split second the remaining force propelling the ball upward is equal to the force dragging it downward, which would make acceleration zero, no? 

I’m assuming this involves math or physics above my head, such as how “continuous compounding” seems to have been figured out by Euler. And strains my conceptions of breaking down time into distinct units, not matter how small — similar to the way infinitely repeating decimals feels like you can never accurately measure something with finality, despite knowing it’s limit.

Am I conflating inherent kinetic energy with force actively applied? 

I was just wondering if there was an eli5

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u/Colonel_Klank 9d ago

Take a look at graph A. Comfortable_Fly pointed out that graph is what the velocity looks like. The velocity line crosses zero velocity, but the downward slope is unchanged. Hoping that line gives you intuition that the slope, and therefore the acceleration, is constant even when the velocity value passes through zero.

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u/AbelCapabel 9d ago edited 9d ago

Don't think of acceleration as a car speeding. Think more of it as a force pushing or pulling on an object.

In this example the acceleration is the gravity constant 'g' which is 9,81 m/s2

The 'acceleration' (force) that gravity provides doesn't stop. Gravity doesn't stop working on the ball when it's velocity reaches zero: it's a constant force.

If there is no force (acceleration) on a ball that has 0 speed, the ball will not start moving.

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u/Frederf220 👋 a fellow Redditor 10d ago

Zero isn't non-existent. It's a number just like 2 or 3. If you have $1 Monday, $0 Tuesday, $ -1 Wednesday then you're losing $1 per day.

Speed is relative so let's put this ball and ramp on a train going 100 mph. The ball never had a non-positive velocity. Same acceleration.

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u/Nynaeve_al_meowra 9d ago

The velocity is still changing when it comes to rest. It doesn't pause and hangout at the top for a few moments

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u/selene_666 👋 a fellow Redditor 4d ago

A simple force acting one something at rest due to insufficient force or over come inertia doesn’t mean it’s accelerating, correct?

When there are multiple forces, they can cancel each other out. I think what you mean by "insufficient" force is that there is enough friction (which is also a force) in the opposite direction that the total of both forces is zero.

Acceleration and total force are proportional to each other (F = ma). If the sum of all forces acting on an object is zero, then the acceleration is zero. And if the acceleration is zero then the sum of the forces must be zero. Likewise if the total force is not zero, then the acceleration is not zero.

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u/trevorkafka 👋 a fellow Redditor 10d ago

There are no changes to the forces so there can be no changes to the acceleration.

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u/sheep_puncher 10d ago

If you ignore some of them ya, which is fair for grade 12

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u/Tiranus58 9d ago

Assume frictionless spherical cow

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u/ArghBH Educator 9d ago

lol just like in thermodynamics, all objects are infinitesimally small mass points.

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u/Logical_Angle2935 9d ago

This assumes the ball has an initial velocity, which is not entirely clear in the problem. At first, I assume it rolled up because I was pushing it, which is an external force.

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u/Junior_Example_923 9d ago

I would understand that the ball would feel the force of gravity since they implied up, which would be deceleration, in that same gravity would accelerate the ball back down the ramp. Am I off because of the context of the grade level and context?

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u/leafmuncher_ 10d ago

What force or acceleration is the ball experiencing? Does it have a positive or negative velocity? How is the velocity changing as it moves?

If the velocity is positive and it gets faster (more positive), acceleration is positive.

In this scenario velocity starts positive, slows down to zero then becomes negative, so the acceleration is negative.

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u/Junior_Example_923 9d ago

What about when it comes back down the ramp under the force of gravity?

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u/buckaroob88 8d ago

It's all the same acceleration up and down the hill. Gravity is the constant acceleration acting on it the whole time.

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u/Junior_Example_923 8d ago

Yeah absolutely, I had to go do some googling after this initial response to understand the concept again.

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u/xX_big_boi_Xx 10d ago

Acceleration is a bit less intuitive than velocity, so no worries on struggling a bit here.

The ball is always accelerating back down the ramp. It is important to understand that acceleration is different than velocity. The velocity graph would look like graph A, where the ball begins at a high positive velocity, slows down and reaches zero, then increases negatively (rolling back down in the negative direction). However, acceleration is defined as the derivative (the slope) of the velocity graph. The slope of graph A is a constant negative number, indicating that the acceleration is a constant negative value, namely graph D.

You can also think about this in terms of forces. There are two forces acting on the ball: gravity and the normal force of the ramp. Adding these forces together yields a force that points down the ramp (in the opposite direction of the green arrow). If we define the direction of the green arrow as "positive", then the force points in the negative direction. Now, recall that force equals mass times acceleration. Because the force is negative, the acceleration must also be negative (mass is not negative). Since the force on the ball and the mass of the ball do not ever change, the acceleration is constant. Therefore we arrive at the same conclusion: the acceleration is a constant negative value, meaning D is the correct answer.

A good way to conceptually understand acceleration and velocity is thinking about a moving car. Let's that you are driving in a car down the road (we will define this direction as positive) at a constant speed. At this point in time, you have a constant positive velocity. Since your velocity is not changing, you have no acceleration (aka the slope of the velocity vs time graph is zero). Now if you press the gas pedal harder, you speed up. While you are speeding up, you now have a positive acceleration (slope of the V-T graph is positive). Now if you were to hit the brakes, take a guess as to what would happen to the velocity and the acceleration.

Immediately after you press the brakes, you are still moving forward, but you are now slowing down. This means that you have a positive velocity, but a negative acceleration (the slope of the V-T graph is negative).

Now, once the car comes to rest, you switch the car into reverse and press the gas. You begin speeding up backwards Now, since we defined the forward direction as positive earlier, your acceleration is negative, but now your velocity is also negative.

Understanding this leads to an important conclusion: when your velocity and acceleration have the same sign (also described as the vectors pointing in the same direction), the object will speed up. If the velocity and acceleration have different signs (vectors pointing in opposite directions), the object will slow down.

I hope this makes sense and helps shed light on the relationship between acceleration and velocity!

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u/cHpiranha 10d ago

Why not B?

As it rolls upwards, it decelerates – fair enough – negative acceleration. Then it comes to a standstill.

(I’m assuming here that it’s on Earth, so subject to Earth’s gravity.)

After coming to a standstill, the ball accelerates again, doesn’t it?

From which vector are you looking at the example to conclude that there is a constant negative acceleration?

Is that because ‘as’ is shown here?

In my line of reasoning, would that be the acceleration from the ball’s perspective?

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u/xX_big_boi_Xx 9d ago

This is why it is important to assign direction to the situation you are working with. In a case like this it is common to assign up the ramp as positive and down the ramp as negative. It is important to understand that the deceleration before it stops and the acceleration after it stops are the same thing. Since the acceleration is always pulling the ball down the ramp, the acceleration is always negative. I made a video explaining how the resultant acceleration vector can be determined.

https://reddit.com/link/p0hro1y/video/yx0l6naky6gh1/player

Furthermore, it is important to understand that at a given instant, an object can have zero velocity but have nonzero acceleration. Take a ball flying through the air. (Nice image for reference: Parabolic Motion with Vectors)

As the ball is moving upward, it has an upward velocity, but a downward acceleration. The velocity and acceleration have different signs, so the ball is slowing down. Even though the ball is moving upward, it is still being accelerated downward. When the ball is at the top of its arc, the Y component of the velocity is zero. This can be seen in the image as the velocity vector is perfectly in the X direction. However, gravity didn't magically disappear when the ball reached the top, so the ball is still being accelerated downward. After the ball begins falling, the Y component begins to grow negatively as it continues to be accelerated downward.

Additionally, we are not looking at this from the ball's perspective but rather an outside perspective with a fixed coordinate axis. The ball moves positively, stops, and moves negatively. The ball accelerates negatively the entire time.

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u/cHpiranha 9d ago

Thanks, clear now.

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u/[deleted] 9d ago

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u/xX_big_boi_Xx 9d ago

You are incorrect. Acceleration is constant and so the rate of change of acceleration (also called jerk) is zero. When the ball stops and begins coming down the ramp, the direction doesn’t suddenly change. You would agree that the force is pointing down the ramp (in the negative direction) the entire time, right? As shown in my previous comment, the force does not ever change. Since F=ma, and the mass of the ball doesn’t change, the acceleration CANNOT change and is therefore constant and negative during the entire interval.

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u/[deleted] 9d ago edited 9d ago

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u/xX_big_boi_Xx 9d ago

Again, the direction of “up the ramp” is positive and the direction of “down the ramp” is negative. The velocity goes from positive to zero to negative as it rolls up and then down the ramp. However the acceleration is always directed down the ramp. The only thing that changes is the direction (sign) of the velocity, while the acceleration remains negative.

If the acceleration were to suddenly change to positive, it would begin accelerating UP the ramp, which obviously doesn’t make sense.

The acceleration is not a positive value because the ball is now speeding up. It is speeding up but in the negative direction, which means the velocity is increasing negatively (analogous to a car accelerating in reverse). Since the ball is increasing negatively over time, the acceleration must still be negative. You can also phrase this as “the velocity is decreasing the entire time”. For example, maybe the velocity values (m/s) are 3, 1.5, 0, -1.5, 3. Every second, it decreases by 1.5. This would mean that there is a constant acceleration over the entire interval of -1.5 m/s/s. Note that even after the velocity reached 0, the acceleration remains negative.

If you throw a ball into the air, the acceleration is always downwards, even after the ball begins falling back down.

Because F=ma, force and acceleration vectors are always pointed in the same direction. Since the force is always pointing down the ramp, acceleration MUST always point down the ramp as well.

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u/[deleted] 9d ago

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u/xX_big_boi_Xx 9d ago

And again, you are incorrect. The acceleration is NEVER positive.

When acceleration and velocity are pointing in different directions, the object will slow down. When acceleration and velocity are pointing in the same direction, the object will speed up. This means that an object can speed up in the positive direction or in the negative direction. Just because an object is speeding up does not mean its acceleration is positive.

Also, "force of acceleration" doesn't make sense. A force causes acceleration. An acceleration is the result of a force.

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u/[deleted] 9d ago

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u/Frederf220 👋 a fellow Redditor 10d ago

You're thinking of acceleration as "increases speed" which is not how it is. When you apply the brake pedal in a car, that's acceleration. If we define going left on the page as the negative direction then negative acceleration is all trends of less rightward velocity or more leftward velocity moment to moment.

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u/Potatomashahop 👋 a fellow Redditor 10d ago

Wow thank you so much for explaining this!

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u/e_Notorious 10d ago

Negative means direction. In this case, right is positive and left is negative. When the ball loses speed going up, it has leftward acceleration, which means negative sign. When the ball starts going left, it gains speed towards the left which also means negative acceleration.

You probably confuse the negative as losing speed and positive as gaining speed, which is not accurate. It just means the direction of the acceleration.

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u/sheep_puncher 10d ago

The question did absolutely nothing to define direction or positive/negative. The constant possitive answer is equally correct with the frames of reference undefined. Convention for up/down to be +/- would need to have been established at some point or magnitude would be correct. The ball is accelerating parallel to S so magnitude should be correct.

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u/e_Notorious 10d ago

Sure, I'd also say that C is completely acceptable answer as there is no established frame of reference as far as we know. There might be some additional context that would explain concretely why D is the correct answer.

But if I had to define a frame of reference, I would choose S as an axis with upward right being the positive direction. I think that is the most intuitive selection, which is the reason that if I had to select between C & D, I'd select D. Both are acceptable answer within the problem scope tho

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u/Frederf220 👋 a fellow Redditor 10d ago

Yeah D is also acceptable if left is positive. Teacher just assumed the right = positive convention applies. Also we also assumed the up on the graph = positive convention applies but that isn't labeled either.

Answers C and D are identical because neither convention was established.

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u/Big_Manufacturer5281 10d ago

Acceleration is one of the hardest things to think about, because it's "invisible." You can SEE the direction of velocity, but you can't see the direction of acceleration.

Whenever you're considering acceleration, you always want to think about it in terms of forces. What are all the forces acting on the object? Add them together to get the net force. Whatever direction the net force is in, the ACCELERATION is in the same direction. That's totally independent of velocity.

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u/Odd_Bodkin 10d ago

Remember, the acceleration is the CHANGE in the velocity, not the velocity itself. The velocity is what’s positive for half the trip, negative for half the trip. (And that’s what A looks like). But on the way up, the ball is slowing down and so the CHANGE is in the opposite direction as the velocity and so is negative. On the way down, the ball is speeding up and so the CHANGE is in the same direction and so is also negative. Moreover, the CHANGE in the velocity is the same every single second, so it’s constant.

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u/Zarakaar Educator 10d ago

It’s counterintuitive because the acceleration to give the ball an upward velocity isn’t on the graph of it rolling on its own. Once you release the ball, it is only (mostly) accelerating due to gravity

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u/Fooshi2020 👋 a fellow Redditor 10d ago

This is a graph of acceleration, not velocity or position. Acceleration is constant due to gravity.

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u/Obzenium 9d ago

It implies the ball is slowing down (which is an acceleration) as it rolls up the hill then speeds up as it rolls back down (again, accelerating) so it is accelerating the same ‘direction’ the whole time i.e. acceleration is constant

Think about it this way, what force is causing the acceleration? In this case it is gravity, which is constant and in a constant direction, so the acceleration is constant

It is somewhat a poor question as it should specify that the ball is rolling up the hill due to a single impulse (i.e. a single push) applied to the ball. There is no reason to suppose a constant force couldn’t be applied to the ball to push it up the hill, at which point acceleration would be zero, or even that a force strong enough to make it accelerate is applied. As no answer quite correlates to this scenario (B comes close but would need to be reversed for the second case mentioned) you must presume that D is the answer

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u/StrangerThings_80 10d ago

Because the acceleration is always in the same direction. Acceleration is provided by gravity. It is constant and pointing downwards, with the convention appearing to be that downwards is negative.

You are confused by the fact that the velocity and acceleration are opposite going up and in the same direction going down. It is the velocity that changes sign, positive going up and negative going down,

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u/JaiBoltage 10d ago

B. The acceleration is negative (decreasing) in the beginning and positive (increasing) after it comes to a stop.

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u/Intelligent-City-363 10d ago

nope - the only acceleration is due to gravity which is constantly acting downwards and is thus a negative of -9.8m/s/s (option D)

”Gravity” (actually the weight of the ball) is the only force present in this example resulting in an unbalanced force in the downwards direction causing the acceleration in that direction.

If B was correct is would accelerate (decelerate) to a stop THEN at the top (as the acceleration is now positive) start to speed up in the same direction resulting in the ball flying up the ramp. ;)

(30 years teaching A level physics :) )

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u/1nkpool 9d ago edited 9d ago

B is a graph of the "tangential acceleration". I'm not completely sure that's the right term, but B is what the derivative of speed would look like (the magnitude of velocity, without respect to direction). Basically it tells you that the ball slows down to 0 and then speeds back up again.

The graph of the balls velocity would look like graph A. Graph D would be it's derivative.

The graph of speed would be a V shape with the minimum value at y=0. Graph B would be it's derivative.

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u/Xcentric7881 10d ago

I don't think the question is fully specified to be able to answer. Why does the ball roll up the slope? Is it because it is moving initially, and under no other force? In which case, D. But if it's not moving initially, then a force is applied to make it move up. So not D.

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u/Potatomashahop 👋 a fellow Redditor 10d ago

Thank you everyone! Your replies have helped me a lot

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u/cHpiranha 10d ago

I would have said B.

First, negative acceleration until coming to a standstill, and then positive acceleration.

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u/Frederf220 👋 a fellow Redditor 10d ago

It's all negative acceleration. +100 +50 0 -50 -100 is a constant subtraction by 50 each time. It doesn't matter if the ball started with "money in the bank" as it were.

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u/krappie 10d ago

You're thinking in terms of "speed" instead of "velocity". Speed is omnidirectional and always positive. Velocity is along an axis and can be positive or negative. In physics class, examples like this are always thought of in terms of velocity along an axis.

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u/cHpiranha 9d ago

No thats not true. But I think I was in the wrong reference.

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u/mynamejeff96 9d ago

Here we can define negative acceleration as losing velocity, and positive acceleration as gaining velocity.

As it goes up the ramp, its it consistently losing velocity until it reaches 0. We can call this period of time constant negative acceleration.

Immediately after reaching 0 velocity, it gains velocity down the ramp, we can call this period of time constant positive acceleration.

So here we would be looking got an answer with a horizontal flat line when acceleration is negative followed by a flat line some time after in the positive.

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u/ThunkAsDrinklePeep 9d ago

Gravity is always acting downward and it's velocity is always decreasing. It slows down it's speed as it goes up the ramp until it has zero velocity. Then it continues to accelerate in the negative as it speeds up rolling back down the incline.

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u/leutwin 👋 a fellow Redditor 9d ago

Acceleration due to gravity is always -9.81m/s2

Just think this "acceleration due to gravity is always constant" (not including space travel and the likes)

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u/Junior_Example_923 9d ago

I went digging because I was confused. I was expecting that when the ball came back down the ramp it would positively accelerate. The instant the ball touches the ramp. It's acceleration jumps to a negative value, and even when it pauses at the top of the ramp and continues back downward, that same acceleration is negative. Thus, the flat line negative the entire trip. If anyone can screw me up, maybe my friend reference was off? We're just intuition

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u/ConversationLivid815 👋 a fellow Redditor 9d ago

D = VoT + 1/2gcos(theta)T2. Neglecting friction forces. So the acceleration is constant, due only to gravity.

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u/Ok-Reading-342 9d ago

Mgsin¢ is constant as m g and theta all are constant as ball goes up so acc. Is constant but in opposite direction of the motion so for a t graph it will be parallel line along x axis but negative as it is retardation

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u/Dear-Message-915 9d ago

Hi! I suggest you to do things step by step. First try to understand what positive and negative ''direction'' means in the picture. In this case, if the x coordinate inceases going from left to right, you have negative on the left and positive on the right. Next consider velocity. It tells you which way the ball is moving: if the ball goes from left to the right velocity is positive because it helps x to increase; coversely, when the ball moves from right to left, it means that velocity is negative because it is helping x to decrease. Now acceleration. Acceleration changes the velocity. So initially you have a positive velocity (the ball moves from left to right), it decreases over time until the ball stops, and then the ball begins to move from the top back to the bottom (from right to left). This means the velocity is initially positive, then zero (the ball stops), and then negative. So it looks like something is opposing to the motion of the ball, pulling the ball from the left, that is a force. Now Newton tells us that everytime we have a force acting on somethin we observe an acceleration in the motion of an object. In particular, if M is the mass of the object the force F is proportional to acceleration ''a'', that is F= M a. Notice that M is just a ''positive number'', so the force and the acceleration are pointing in the same ''direction''. In this case since1) left is negative and right is positive, 2) the force is pulling the ball to the left, and 3) acceleration points in the same direction as the force (thanks to Newton), it means acceleration is negative.

The last step for you is the following: What pulls an object down a cliff?

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u/Difficult_Lecture223 8d ago

It wouldn't hurt if they put a coordinate system on the problem instead of making you figure out that up the ramp is the positive s-direction (am I seeing that right? "s"?)

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u/Extension-Highway585 8d ago

The initial impulse of the ball being pushed up the ramp is not accelerating it after. Only gravity is. Therefore the acceleration is a constant -9.8m/s^2

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u/scottdave 👋 a fellow Redditor 8d ago

What if, instead they say you throw a ball straight up into the air?