# Homework Help: Acceleration Question

1. Sep 6, 2004

### jtagtp

An Automobile starting from a rest at t=0 undergoes constant acceleration on a staight line. It is observed to pass two marks separated by 64m, first at t=8 and the second at t=12s. What is the value of the acceleration?

I know how to find the Vel. from this (64m/4s) but i dont understand how im able to find the acceleration?

The answer in the book is 1.6 m/s^2 but i what to understand how to do it. Thank You

J

2. Sep 6, 2004

### arildno

1.Your expression for the (instantaneous) velocity is totally wrong; you're mixing up with the average velocity.

How is "distance travelled in time t" related to constant acceleration?

3. Welcome to PF!

3. Sep 6, 2004

### 97gtpacecar

Man im so lost. Ok i relized i used the average velocity, but what i need to do is use the instantaneous velocity? I never did understand inst. vel?

Thanks for welcoming me, i loved physics in high school but so far this college physics is

4. Sep 6, 2004

### arildno

OK, let's take it slow&easy:
What formulas do you know which is about constant acceleration?

5. Sep 6, 2004

### 97gtpacecar

constant acceleration =(v(t2)-v(t1))/(t2-t1) = chance in v over the change in time.

6. Sep 6, 2004

### HallsofIvy

Do you know d= (1/2)a t2 for constant acceleration starting from rest?

7. Sep 6, 2004

### 97gtpacecar

So i guess i can try and use the formula and switch it around. a=2d/(t^2)

8. Sep 6, 2004

### 97gtpacecar

ummm but the problems is that this in not from a dead stop.

9. Sep 6, 2004

### needhelpperson

Either the answer in the book is wrong, or the question is flawed.

we know that vi = 0

so vf/2 = 64/8 = 16m/s

a = 16m/s/8s = 2m/s^2. If the acceleration is constant then it should be 2m/s^2 the whole time. However, i don't know where the 1.6m/s^2 came from, unless it didn't start from rest....

10. Sep 6, 2004

### 97gtpacecar

Well it starts from a rest but what we are looking is already moving. It start at 0 but what we are looking at is the section from 8t to 12t

11. Sep 6, 2004