# Instantaneous Velocity and Acceleration

## Homework Statement

A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following
a. The average velocity of the car from t=0s to t=10s
b. The instantaneous velocity of the car at t=10s
c. The average acceleration of the car at t=10s
d. The instantaneous acceleration of the car at t=10.s

Derivatives

## The Attempt at a Solution

in a.) I integrated the equation to get the position function
x(t)=(1/2)(7.20m/s^2)(t^2)+(1/3)(-0.720m/s^3)(t^3)+C
then I substitute t = 10 s
I got 120 m/s but the answer should be -72 m/s <- So I need help here

b.) I derive the x(t) to return the original equation then I substitute t = 10 s right?
c.) I'll simply derive the original equation (7.20m/s^2)+2(-0.720)(10) = -7.2 m/s^2
d.) Use the original equation then plug in t = 10 s ? <- Help

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SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following
a. The average velocity of the car from t=0s to t=10s
b. The instantaneous velocity of the car at t=10s
c. The average acceleration of the car at t=10s
d. The instantaneous acceleration of the car at t=10.s

Derivatives

## The Attempt at a Solution

in a.) I integrated the equation to get the position function
x(t)=(1/2)(7.20m/s^2)(t^2)+(1/3)(-0.720m/s^3)(t^3)+C
then I substitute t = 10 s
I got 120 m/s but the answer should be -72 m/s <- So I need help here
If x(t) is the position of the car, why are the units m/s? That's your first problem.

Exactly which question does -72 m/s answer?

Chestermiller
Mentor

## Homework Statement

A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following
a. The average velocity of the car from t=0s to t=10s
b. The instantaneous velocity of the car at t=10s
c. The average acceleration of the car at t=10s
d. The instantaneous acceleration of the car at t=10.s

Derivatives

## The Attempt at a Solution

in a.) I integrated the equation to get the position function
x(t)=(1/2)(7.20m/s^2)(t^2)+(1/3)(-0.720m/s^3)(t^3)+C
then I substitute t = 10 s
I got 120 m/s but the answer should be -72 m/s <- So I need help here
Where did the -72 m/s come from? I calculated that the distance covered over 10 sec was 120 m. So the average velocity should be 12 m/s. During the 10 sec, the instantaneous velocity is never zero, so the average certainly shouldn't be negative.

Chet

Where did the -72 m/s come from? I calculated that the distance covered over 10 sec was 120 m. So the average velocity should be 12 m/s. During the 10 sec, the instantaneous velocity is never zero, so the average certainly shouldn't be negative.

Chet
I borrowed my friend's notebook because I studying about kinematic then I saw this example and resolved it. The answer on a letter A is -72(on his notebook) but I got 120 soooooo I posted in this forum to confirm which is the right answer

If x(t) is the position of the car, why are the units m/s? That's your first problem.

Exactly which question does -72 m/s answer?
oooops I meant to say 120 m

Chestermiller
Mentor
I borrowed my friend's notebook because I studying about kinematic then I saw this example and resolved it. The answer on a letter A is -72(on his notebook) but I got 120 soooooo I posted in this forum to confirm which is the right answer
So, from what you know now, was his answer right or not?

So, from what you know now, was his answer right or not?
First I was frustrated that I can't get -72 m which consuming all my time so I decided to post my silly question (sorry guys I have to do it time is gold right?)

but I conclude that maybe he miscalculate his problem or something
but it is confirm the answer in a.) is 120 m? I always get 120 m. Sooooooo I hope I'm right. This problem is already wasted 1 hour in my review session.

Chestermiller
Mentor
First I was frustrated that I can't get -72 m which consuming all my time so I decided to post my silly question (sorry guys I have to do it time is gold right?)

but I conclude that maybe he miscalculate his problem or something
but it is confirm the answer in a.) is 120 m? I always get 120 m. Sooooooo I hope I'm right. This problem is already wasted 1 hour in my review session.
What is the definition of average velocity? Does velocity have units of meters?

Chet

What is the definition of average velocity? Does velocity have units of meters?

Chet
I meant to say when you're getting x(t) the units should m or meters right? So I answered 120 m. Then I use average velocity = ΔX/ΔT soooo 120 m / 10 s = 12 m/s right?
This is for letter A

Chestermiller
Mentor
I meant to say when you're getting x(t) the units should m or meters right? So I answered 120 m. Then I use average velocity = ΔX/ΔT soooo 120 m / 10 s = 12 m/s right?
This is for letter A
Correct.

Chet

Correct.

Chet
for letter B finding the in instantaneous velocity. I don't need to derive it because it is already in v(t)
I'll just plug t = 10 s into v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2

I got 0 m/s hmmm I find it odd to get 0 in instantaneous velocity

Chestermiller
Mentor
for letter B finding the in instantaneous velocity. I don't need to derive it because it is already in v(t)
I'll just plug t = 10 s into v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2

I got 0 m/s hmmm I find it odd to get 0 in instantaneous velocity
Have you ever heard of speeding up, then slowing down, and finally stopping.

Chet

Have you ever heard of speeding up, then slowing down, and finally stopping.

Chet
Oh okay that make sense to prove that it I can get 0 m/s thank you!!!

Very good job. You were very close on a.
..The integration was correct.
..Just a minor math error

## Homework Statement

A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following
a. The average velocity of the car from t=0s to t=10s

## The Attempt at a Solution

in a.) I integrated the equation to get the position function
x(t)=(1/2)(7.20m/s^2)(t^2)+(1/3)(-0.720m/s^3)(t^3)+C
then I substitute t = 10 s
I got 120 m/s but the answer should be -72 m/s <- So I need help here
1) Remember, you are calculating average velocity for the 10 second period. Since the initial velocity is 0 m/s. The final velocity is 0 m/s. And all velocities in between are greater than zero there is no way the answer can be a negative number.

2) You have to dispose of the C by applying the equation when t=0. Naturally C= 0 so you can dispense with it.

3) You calculated x(t) at 10 seconds to be 120 Meters (not m/s). So you got the distance traveled in ten seconds correct.

4) You are going for average velocity. so that is the distance traveled divided by the time. v avg = [x(10s) - x (0s) ] / [t(10s) - t(0s)]
V avg = 120 m / 10 s = 12 m/s

Question: Now how do you check the answer?

What is the difference between AVERAGE acceleration and INSTANTANEOUS acceleration.
{Perhaps c. should be from t=0s to t=10s}

c. The average acceleration of the car at t=10s
d. The instantaneous acceleration of the car at t=10.s

Practicing Latex

Velocity
$$v(t)=(7.20m/s^2) \times t-(0.720m/s^3) \times t^2$$

Distance traveled is the area under the curve.
So he needed to integrate the velocity equation, which he did correctly.
$$\int v(t) \, dt = x(t) = \frac {7.20m/s^2 \times t^2} {2} \quad + \quad \frac {-0.720m/s^3 \times t^3} {3} \quad+ \quad C$$

And Acceleration is the slope of the velocity line.
So he needed to derive the velocity equation, which he did correctly.
$$\frac {\partial v(t)} {\partial t}= a(t) = 7.20m/s^2 \quad + \quad 2 \times (-0.720)m/s^3 \times t \quad$$