(adsbygoogle = window.adsbygoogle || []).push({}); 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

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

3. The attempt at a solution

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, but thats not the anwser :s

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

2. Relevant equations

3. The attempt at a solution

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# Kinematics problem that i cant seem to resolve

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