Question

a bike travelling at the velocity of 15m/s is stopped by applying brakes with uniform retardation of 1m/s^2. calculate distance before bike coming to rest and the time taken​

Answer 1

Explanation:

a= -1 m/s^2 (retardation)

v^2 = u^2 + 2*a*s

v = 0

2*1*s = 15^2

s = 225/2 = 112.5 m

v =u + a*t

t = 15/1 = 15 sec

Answer 2

Answer:

112.5 m & 15 seconds

Explanation:

Stopping distance = (v² - u²)/(2a)

= (0² - 15²) / (2 × -1 m/s²)

= (225/2) m

= 112.5 m

Time taken to come to rest

Use equation of motion

• v = u + at

T = (v - u)/a

= (0 m/s - 15 m/s) / (-1 m/s²)

= 15 s