Question

A car starts from rest and moves with constant acceleration of 4 m/s2 for 30 seconds. Then the brakes are applied and the car comes to rest in another 60 seconds. Find[5] (i) Maximum velocity attained by car(ii) Magnitude of retardation of car(iii) Total distance covered by car

Answer 1

Answer:

120 m/s, -2 m/s2, 5400m

Explanation:

In first 30 second car's velocity is increasing and after that it is decreasing. so maximum velocity is at 30 seconds. Call it V30.

V30 = V0 + a x t

V30 = 0 + 4 x 30

V30 = 120 m/s2

Distance covered in 30 seconds, d30 = $\frac{(V30)^{2} - {0}^2}{2*a}$ = $\frac{120^2}{2*4}$ = 1800 m

Now in next 60 seconds car's velocity comes down from V30= 120 to 0 m/s.

So, using above formula,

0 = V30 + a x 60

0 = 120 + 60a

a= -2 m/s2 . so retardation magnitude = 2 m/s2

distance covered in 60 seconds= d60 =  $\frac{(V60)^{2} - (V30)^2}{2*a}$ = $\frac{(0)^{2} - (120)^2}{2*(-2)}$

d60= 3600 m

Total distance covered = d30+d60 = 1800+3600 =5400 m