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

Answer:

464.69 m

Explanation:

First car

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}2.84t^2\\\Rightarrow s=1.42t^2$

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

$1.42t^2=25.7t\\\Rightarrow t=\frac{25.7}{1.42}\\\Rightarrow t=18.09\ s$

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