# Acceleration-time graph: average distance and average force

• moenste
In summary: The correct labeling should be 0 at the 5 second mark instead of 5.With that in mind, let's revisit the questions and the solutions:(a) the average distance traveled in 15 s:In the first 5 seconds, the object has a constant acceleration of 2 m/s2. Using the formula s = (1/2) a t2, we get s = (1/2) * 2 * 52 = 25 m. In the next 10 seconds, the object has a constant velocity of 10 m/s. Using the formula s = v*t, we get s = 10 * 10 = 100 m.Therefore, the total distance traveled
moenste

## Homework Statement

The acceleration-time graph above is drawn for a body which starts from rest and moves in a straight line. The body is of mass 10 kg.

Use the graph to find:
(a) the average distance traveled in 15 s,
(b) the average force acting over the whole 15 s period.

Answers: (a) 125 m, (b) 6.66667 N / 6.2/3 N

-

## The Attempt at a Solution

(a) v = 2 m s-2 * 5 s = 10 m s
s = 10 m s * 5 s = 50 m

or

s (t) = 1/2 a t2
s = 1/2 * 2 * 152 - 1/2 * 2 * 102 = 125 m

In both cases I assumed that it's 5 and not 0 in the Time/s graph.

But I am completely not sure about the solutions and also if the acceleration was only 5 seconds why can't I get 125 m using just 1/2 * 2 * 52 = 25 m...

moenste said:

## Homework Statement

The acceleration-time graph above is drawn for a body which starts from rest and moves in a straight line. The body is of mass 10 kg.

Use the graph to find:
(a) the average distance traveled in 15 s,
(b) the average force acting over the whole 15 s period.

Answers: (a) 125 m, (b) 6.66667 N / 6.2/3 N

-

## The Attempt at a Solution

(a) v = 2 m s-2 * 5 s = 10 m s
s = 10 m s * 5 s = 50 m

or

s (t) = 1/2 a t2
s = 1/2 * 2 * 152 - 1/2 * 2 * 102 = 125 m

In both cases I assumed that it's 5 and not 0 in the Time/s graph.

But I am completely not sure about the solutions and also if the acceleration was only 5 seconds why can't I get 125 m using just 1/2 * 2 * 52 = 25 m...

With regard to method (a), is the velocity of the object = 10 m/s for the whole 5 second period, or just after 5 seconds has elapsed?

Same with calculating the distance traveled. Does the body accelerate at 2 m/s2 for the entire 15 second interval or just the first 5 seconds? What happens to the velocity of the body after the first 5 seconds has elapsed?

Remember, these formulas v = at and s = (1/2) a t2 only apply when the acceleration is a constant. If you have two different values of constant acceleration, you have to apply the formulas to the time intervals which have a constant acceleration separately if the acceleration is a different, though constant, value.

moenste
SteamKing said:
With regard to method (a), is the velocity of the object = 10 m/s for the whole 5 second period, or just after 5 seconds has elapsed?

Same with calculating the distance traveled. Does the body accelerate at 2 m/s2 for the entire 15 second interval or just the first 5 seconds? What happens to the velocity of the body after the first 5 seconds has elapsed?

Remember, these formulas v = at and s = (1/2) a t2 only apply when the acceleration is a constant. If you have two different values of constant acceleration, you have to apply the formulas to the time intervals which have a constant acceleration separately if the acceleration is a different, though constant, value.
I think the body gains 10 ms velocity at 5 seconds.

It is shown in the graph that the body accelerates for 5 seconds. After than the velocity and acceleration are 0.

How can I apply 0 acceleration? s = 1/2 a t2 = 1/2 * 2 ms * 52 s = 25 meters. The answer is 125 m in the book. Where am I wrong?

moenste said:
I think the body gains 10 ms velocity at 5 seconds.

It is shown in the graph that the body accelerates for 5 seconds. After than the velocity and acceleration are 0.

I agree that the graph shows that the acceleration of the body is zero after two seconds. But you've lost me on why the velocity is also zero after two seconds.

Remember Newton's Fourth Law of Motion: Velocity ain't Acceleration, baby!

How can I apply 0 acceleration? s = 1/2 a t2 = 1/2 * 2 ms * 52 s = 25 meters. The answer is 125 m in the book. Where am I wrong?
See comment above.

moenste
SteamKing said:
See comment above.
So
s (t) = 1/2 a t2
s = 1/2 * 2 * 152 - 1/2 * 2 * 102 = 125 m
Right?

moenste said:
So
s (t) = 1/2 a t2
s = 1/2 * 2 * 152 - 1/2 * 2 * 102 = 125 m
Right?
No, that's not right. It gives you the right answer, but for the wrong reasons.

Look, let's take this in stages. How far does the object travel in the first two seconds, from the origin?

What is the acceleration of the body after two seconds?

What is the velocity of the body after two seconds?

What is the velocity of the body after three seconds? How far has the body traveled, from the origin?

What is the velocity of the body after four seconds? How far has the body traveled, from the origin?

What is the velocity of the body after five seconds? How far has the body traveled, from the origin?

moenste
The graph I see has an acceleration of 2m/s2 prior to time, t = 0, and it looks as if the object has that acceleration for the 5 second interval immediately prior to t = 0. Is the object at rest at t = -5s ? The acceleration is zero from t = 0 to 15s.

Which 15 second interval do the questions refer to?

Ignore the previous part of this post .

I see that the time axis is simply mislabeled . There is a zero at the 5 second mark instead of a 5 .

Last edited:
SammyS said:
The graph I see has an acceleration of 2m/s2 prior to time, t = 0, and it looks as if the object has that acceleration for the 5 second interval immediately prior to t = 0. Is the object at rest at t = -5s ? The acceleration is zero from t = 0 to 15s.

Which 15 second interval do the questions refer to?
The thinking is that the time scale is graduated into 5-second increments, and the first increment is labeled "0" instead of "5". IOW, the axis showing magnitude of acceleration is located at t = 0 sec.

SteamKing said:
How far does the object travel in the first two seconds, from the origin?
If 0 seconds on the vertical axis is 5 seconds then I think the object travels: s = 1/2 a t2 = 1/2 * 2 m s-2 * 52 seconds = 25 meters. The body traveled 25 meters in the first 5 seconds.

What is the acceleration of the body after two seconds?
After 5 seconds the acceleration is equal to 0 m s-2.

What is the velocity of the body after two seconds? What is the velocity of the body after three seconds? How far has the body traveled, from the origin? What is the velocity of the body after four seconds? How far has the body traveled, from the origin? What is the velocity of the body after five seconds? How far has the body traveled, from the origin?
Hm, should not the velocity be zero after five seconds?
v = u + a t
v2 = u2 + 2 a s
u = velocity when t = 0
v = velocity at time t
a = constant acceleration
s = the distance from the starting point at time t (not necessarily the distance moved)

u = 0 when t = 0
a = 0 m s-2 after 5 seconds
s = 25

v = 0 + 0 m s-2 * 6 seconds = 0
v2 = 02 + 2 * 0 m s-2 * 25 m = 0

Also another formula: a = v / t
v = a t
v = 0 m s-2 * 6 seconds = 0.

And as a result the body has traveled 25 meters in total.

moenste said:
If 0 seconds on the vertical axis is 5 seconds then I think the object travels: s = 1/2 a t2 = 1/2 * 2 m s-2 * 52 seconds = 25 meters. The body traveled 25 meters in the first 5 seconds.After 5 seconds the acceleration is equal to 0 m s-2.
Yes, this is correct.

Hm, should not the velocity be zero after five seconds?

And you still haven't learned Newton's Fourth Law of Motion: Velocity ain't Acceleration, baby!

SammyS and moenste
SteamKing said:

And you still haven't learned Newton's Fourth Law of Motion: Velocity ain't Acceleration, baby!
Newton's First Law: Every body continues in its state of rest or of uniform (unaccelerated) motion in a straight line unless acted on by some external force.

According to the formulas which I could find in the textbook if acceleration is 0 then velocity is equal to 0 aswell. I did show all the calculations.

If velocity is y = 2x, and acceleration is y = 2, where x is time then I have a slope of (0, 0), (1, 2), (2, 4), (3, 6), (4, 8) and (5, 10) coordinates. After that I have acceleration y = 0 and then velocity is y = 0x where x = t. So velocity is zero when acceleration is zero.

moenste said:
Newton's First Law: Every body continues in its state of rest or of uniform (unaccelerated) motion in a straight line unless acted on by some external force.

According to the formulas which I could find in the textbook if acceleration is 0 then velocity is equal to 0 aswell. I did show all the calculations.

You're lost in formulas without understanding what they mean.

Try this explanation of Newton's First Law of Motion:

https://www.grc.nasa.gov/www/K-12/airplane/Newton.html

The key passage is quoted below:
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. The key point here is that if there is no net force acting on an object (if all the external forces cancel each other out) then the object will maintain a constant velocity. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force.

moenste said:
If velocity is y = 2x, and acceleration is y = 2, where x is time then I have a slope of (0, 0), (1, 2), (2, 4), (3, 6), (4, 8) and (5, 10) coordinates. After that I have acceleration y = 0 and then velocity is y = 0x where x = t. So velocity is zero when acceleration is zero.

I'm sorry, but I don't follow this at all. You've leapt to some sort of unjustifiable conclusion here.

If velocity is expressed as v = 2t, then at t = 0, v = 0, but the acceleration is a = 2, and it remains a = 2, regardless of the value of t.

If the acceleration should become zero, then all this means that whatever the velocity is at that time, it remains unchanged; it doesn't drop to zero.

That's Newton's First Law.

moenste
SteamKing said:
If the acceleration should become zero, then all this means that whatever the velocity is at that time, it remains unchanged; it doesn't drop to zero.
I got your point after this sentence.

(i) s1 = 1/2 a t2 = 1/2 * 2 m s-2 * 52 s = 25 m
(ii) v = a t = 2 m s-2 * 5 s = 10 m s-1
(iii) s2 = v t = 10 m s-1 * 10 s = 100 m
(iv) sTotal = s1 + s2 = 25 m + 100 m = 125 m

I think this is right.

For (b) part:
F = (m v) / t
F = (10 kg * 10 m s-1) / 15 s = 6.67 N

Does fit the book answer but I don't quite understand why this is the average force? I mean the object has velocity of 10 m s-1 only after the fifth second, before that it has 0, 2, 4, 6, and 8 m s-1 for the 0, 1, 2, 3, and 4 second respectively. If my solution for (b) is right could you explain why we use the same velocity for the whole period of time? (shouldn't the velocity be different for the first four seconds?)

moenste said:
I got your point after this sentence.

(i) s1 = 1/2 a t2 = 1/2 * 2 m s-2 * 52 s = 25 m
(ii) v = a t = 2 m s-2 * 5 s = 10 m s-1
(iii) s2 = v t = 10 m s-1 * 10 s = 100 m
(iv) sTotal = s1 + s2 = 25 m + 100 m = 125 m

I think this is right.

For (b) part:
F = (m v) / t
F = (10 kg * 10 m s-1) / 15 s = 6.67 N

Does fit the book answer but I don't quite understand why this is the average force? I mean the object has velocity of 10 m s-1 only after the fifth second, before that it has 0, 2, 4, 6, and 8 m s-1 for the 0, 1, 2, 3, and 4 second respectively. If my solution for (b) is right could you explain why we use the same velocity for the whole period of time? (shouldn't the velocity be different for the first four seconds?)

Since the acceleration of the body is constant for 0 ≤ t ≤ 5 sec., then F = ma can be re-written to be F = m * Δv / Δt, where acceleration is the change in velocity Δv divided by the change in time Δt. When 5 ≤ t ≤ 15 sec., the acceleration is zero, which means the net force acting on the body over this interval is also zero, according to F = ma.

So the force for the interval where the body is accelerating is
F = ma = 10 kg * 2 m/s2 = 20 N = (10 kg * 10 m/s) / 5 sec.

The quantity mv = 100 kg-m/s is also known as the momentum of the body during this time.

This force of 20 N is present for 5 sec and is constant during that time, so the area under the force-time curve would be A = 20 N * 5 sec = 100 N-s. After 5 sec, F = 0, so no additional area accumulates under the force-time curve.

The average force = area under the force-time curve, from t = 0 to t = 15 sec., divided by the elapsed time, Δt = 15 sec.

Favg = A / Δt = 100 N-s / 15 s = 6.667 N

moenste

## 1. What is an acceleration-time graph?

An acceleration-time graph is a visual representation of an object's acceleration over time. It shows how an object's velocity changes over a period of time, with acceleration being the rate of change of velocity.

## 2. How do you calculate average distance from an acceleration-time graph?

To calculate average distance from an acceleration-time graph, you can use the formula: average distance = initial velocity x time + 1/2 x acceleration x time^2. This formula takes into account the initial velocity, the time interval, and the acceleration of the object.

## 3. What is the significance of the slope of an acceleration-time graph?

The slope of an acceleration-time graph represents the object's acceleration. A steeper slope indicates a higher acceleration, while a flatter slope indicates a lower acceleration. The slope can also be used to calculate the average acceleration of the object.

## 4. How is average force related to an acceleration-time graph?

Average force is directly related to an acceleration-time graph. According to Newton's Second Law of Motion, force is equal to mass multiplied by acceleration. Therefore, the average force can be calculated by multiplying the mass of the object by its average acceleration.

## 5. How can an acceleration-time graph be used to analyze an object's motion?

An acceleration-time graph can provide valuable information about an object's motion. By analyzing the slope, you can determine the object's acceleration. By calculating the area under the graph, you can determine the object's average distance. Additionally, the graph can also show changes in velocity over time, allowing for a more detailed understanding of the object's motion.

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