# Homework Help: Where and when does the car catch the truck?

1. Sep 5, 2011

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

A truck starts from rest and accelerates at 1 M/S*2 . 4 s later, a car accelerates from rest at the same starting point with an acceleration of 2.7 M/S*2 .

Where and when does the car catch the truck?
2. Relevant equations

3. The attempt at a solution

Last edited: Sep 5, 2011
2. Sep 5, 2011

### lewando

Please clarify what you mean by "acceleration of ."

Please show what you have done so that help can be provided.

3. Sep 5, 2011

i clarified!

4. Sep 5, 2011

### lewando

Thanks, adoule. So what have you tried so far?

5. Sep 5, 2011

i have tried s=ut + (1/2)at^2

In 4 seconds, truck travels a distance of 0 (4) + (1/2) (1) (4)^2 = 0 + 8 = 8

now, when car starts, truck is already 8 m ahead of it.

Suppose they meet at time lapse of t seconds after car starts.
So, car travels 8 m more than the truck in t seconds.
distance travelled by car in t seconds = dist travelld by truck in t seconds + 8

0 (t) + (1/2) (2.7) (t)^2 = 0 ( t) + (1/2) (1) (t)^2 + 8

==> (t)^2 = (160) / 17
t = 3.06

hence, car catches truck at 4 + 3.06 = 7.06 seconds after truck starts.
distance travelled by car = 0 (7.06) + (1/2) (4) ( 160/17) = 18.8 metres

6. Sep 5, 2011

### lewando

I'm with you at this point.
still with you...
The RHS is not correct. You need to take into account the velocity of the truck (not zero!)

7. Sep 5, 2011

ok so i corrected the velocity of the truck it gives me this
0 (t) + (1/2) (2.7) (t)^2 = 4 ( t) + (1/2) (1) (t)^2 + 8
and then it gives me an equation like this
0= 0.85t^2- 4t-8
and the anwser is 6.2192121 but its still nor correct :s