# Simple problem - finding acceleration

1. Aug 25, 2011

### Nat3

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

An object traveling at a constant 100km/hr (27.8 m/s) stops in 0.75m

2. Relevant equations

What was the object's acceleration?

3. The attempt at a solution

$a = \frac{\Delta v}{\Delta t}$

$\Delta t = \frac{0.75 m}{27.8 m/s} = 0.027 s$

$a = \frac{0-27.8 m/s}{0.027 - 0 s} = -1029.63 m/s^2$

The textbook answer is -500 m/s^2. I'm confused :(

2. Aug 25, 2011

### lewando

Your problem is with this line. It would be appropriate if velocity was constant over that distance, but it is not.

3. Aug 25, 2011

### Staff: Mentor

When you calculate your Δt, the velocity is not constant over the time interval. Thus $d \neq \Delta v\; \Delta t$. Rather, the average velocity will be Δv/2, so that
$$d = \frac{\Delta v}{2} \Delta t$$
$$\Delta t = \frac{2 d}{\Delta v}$$

4. Aug 25, 2011

### Nat3

Why is the average velocity Δv/2? What does "d" represent?

5. Aug 25, 2011

### Staff: Mentor

d is the distance traveled (you can call it Δx if you wish). The initial velocity is v, the final velocity is zero. So the Δv is (v - 0). The average is (v + 0)/2, or Δv/2.

6. Aug 25, 2011

### Nat3

Is this average really accurate? For example, if a car drives at 10 m/s for 10 minutes, then slows to a stop in one second, is the average speed really going to be (0+10)/2 ?

7. Aug 25, 2011

### Staff: Mentor

It is for the period of the deceleration; the rest of the trip is not being considered

And it's (10 + 0)/2 in this case.

8. Aug 26, 2011

### Nat3

So this works because we assume the acceleration (deceleration) during the .75m the object is slowing down is constant? If the acceleration changed during that time, then your method of calculating the average velocity would not be accurate?

Thanks :)

9. Aug 26, 2011

Correct.