A truck starts from rest with an acceleration of `1.5 m//s^(2)` car 150 m behind starts from rest with an acceleration of `2 m//s^(2)`. How long will it take before both the truck and car side byside, and how much distance is traveled by each?

Options

A. 2.45 s, 500 m (truck), 650 m (car)
B. 5 s, 450 m (truck), 600 m (car)
C. 24.5 s, 450 m (truck), 600 m (car)
D. 5.3 s, 500 m (truck), 650 m (car)

Explanation

Let x be the distance travelled by the truck when truck and car are side by side. The distance travelled by the car will be (x + 150) as the car is 150 metre behind the truck. Applying the formula
s = ut + (1/2) a `t^(2)`
we have , ` x = 1/2 xx (1.5) t^(2) ` .......(A)
and ` (x + 150 ) = (1/2) xx (B) t^(2) ` ......(B)
Here t is the common time .
From equations (A) and (B)
we have, ` (x + 150)/(x) = (2)/(1.5)`
`{:("Solving we get, x = 450 m ","(truck)"),("and x + 150 = 600 m " ,"(car)"):}`
Substituting the value of x in eq. (A), we get 450 = 1/2 (1.5) `t^(2)`
` therefore t = sqrt((450xx 2)/(1.5)) = sqrt(600) =24.5 ` sec .
Hence correct answer is (C)