Answer
a) Time taken by car to overtake truck = $\frac{20}{\sqrt 7}$ sec
b) Car was $\frac{260}{7}$ metres behind the truck
c) Speeds of when they are abreast are
For car = $\frac{68}{\sqrt 7}$ m/s
For Truck = $\frac{42}{\sqrt 7}$ m/s
Work Step by Step
a) Both vehicles are starting from rest, and when truck moves 60m, the car overtakes it.
So time taken by the car to overtake the truck is same as time taken by the truck to move 60m
Which by equation of motion $s = ut + \frac{1}{2}at^{2}$
where $s = 60m, u = 0, a = 2.1ms^{-2}$
This gives, $60 = \frac{1}{2}(2.1)t^{2}$
=> $t = \frac{20}{\sqrt7}$ sec
b) Distance the car was behind the truck = Distance travelled by the car when it overtakes - Distance moved by the truck in the same time
By $s = ut + \frac{1}{2}at^{2}$
Distance travelled by the car = $ \frac{1}{2}(3.4) (\frac{20}{\sqrt7})^{2}$
= $ \frac{680}{7} metres$
Distance by which the car was behind the truck = $ \frac{680}{7} - 60 = \frac{260}{7} metres$
c) Their speed when they are abreast, or at overtaking time i.e at $t = \frac{20}{\sqrt7}$ sec
Using $ v = u + at$
$u$ i.e initial speed is 0 for both of them
For the car speed = $3.4\frac{20}{\sqrt7}= \frac{68}{\sqrt7}m/s$
For the truck speed = $2.1\frac{20}{\sqrt7}= \frac{42}{\sqrt7}m/s$
