Question

A car starts from rest with an acceleration of 2.84 m/s2 at the instant when a second car moving with a velocity of 25.7 m/s passes it in a parallel line. How far does the first car move before it overtakes the second car?

Answer 1

464.69 m

Step-by-step explanation:

First car

`s=ut+(1)/(2)at^2\\\Rightarrow s=0* t+(1)/(2)2.84t^2\\\Rightarrow s=1.42t^2`

Second car

Distance = Speed × Time

`\text{Distance}=25.7t`

Here, the time taken and the distance traveled will be the same

Equating the two equations

`1.42t^2=25.7t\\\Rightarrow t=(25.7)/(1.42)\\\Rightarrow t=18.09\ s`

So, the first would have to move
`1.42t^2=1.42* 18.09^2=464.69\ m` in order to overtake the second car.