# Instantaneous Velocity and Acceleration

1. Aug 18, 2015

### VaioZ

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Derivatives

3. 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

2. Aug 18, 2015

### SteamKing

Staff Emeritus
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?

3. Aug 18, 2015

### Staff: Mentor

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

4. Aug 18, 2015

### VaioZ

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

5. Aug 18, 2015

### VaioZ

oooops I meant to say 120 m

6. Aug 18, 2015

### Staff: Mentor

So, from what you know now, was his answer right or not?

7. Aug 18, 2015

### VaioZ

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.

8. Aug 18, 2015

### Staff: Mentor

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

Chet

9. Aug 18, 2015

### VaioZ

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

10. Aug 18, 2015

### Staff: Mentor

Correct.

Chet

11. Aug 18, 2015

### VaioZ

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

12. Aug 18, 2015

### Staff: Mentor

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

Chet

13. Aug 18, 2015

### VaioZ

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

14. Aug 18, 2015

### NickAtNight

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

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?

15. Aug 18, 2015

### NickAtNight

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

16. Aug 18, 2015

### NickAtNight

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.
$$\frac {\partial v(t)} {\partial t}= a(t) = 7.20m/s^2 \quad + \quad 2 \times (-0.720)m/s^3 \times t \quad$$