# Equations of Motion Question

1. Oct 28, 2009

### xcompulsion

A car travelling at 40km/h has it's speed reduced to 10km/h in 2.0s. Assuming that the same deceleration would be in effect, find how far the car will travel in coming to rest from a speed of 60km/h?

2. Oct 28, 2009

### hage567

What are you thoughts on how to start this problem?

3. Oct 28, 2009

### xcompulsion

I'm thinking about calculating the acceleration using the givens from the first part, to calculate the second part.

so using the equation,
v2^2 = v1^2 + 2(a)(t)

But what has gotten me really puzzled is the seconds, the hours, and the kilometers.

4. Oct 28, 2009

### hage567

That's the right approach, but that equation is not quite right. That t should be a d (as in distance). Check for another equation.

You will need to convert your quantities so they are all in the same units. I usually work in meters and seconds in these questions. I would start with that before you do anything else.

5. Oct 28, 2009

### xcompulsion

Okay, so after converting all the units to m/s..

first part:
v1 = 40km/h = 11.11m/s
v2 = 10km/h = 2.8m/s
t=2.0s
a = v/t = (2.8m/s - 11.11m/s) / 2.0s = -4.155m/s^2

second part:
v1 = 60km/h = 16.67m/s
v2 = 0km/h = 0m/s
a = -4.155m/s^2
d = ?

v2^2 = v1^2 + 2(a)(d)
0 = (16.67)^2 + 2(-4.155)(d)
0 = 277.9 + (-831)(d)
d = 831 - 277.9
d = 553.1m/s

However, the correct answer to this question is 0.033km..

6. Oct 28, 2009

### hage567

That -831 is wrong, you missed the decimal place!

Also, you didn't solve for d correctly. There should be division involved. Try again.

7. Oct 28, 2009

### xcompulsion

v2^2 = v1^2 + 2(a)(d)
0 = (16.67)^2 + 2(-4.155)(d)
0 = 277.9 + (-8.31)(d)
-277.9 / -8.31 = (-8.31)(d) / (-8.31)
d = 33.44m

so, 33.44/1000 = 0.033km!

thank you so much

8. Oct 28, 2009

### hage567

You're welcome!