Question

Find the velocity of a car which is stopped in 10 seconds by applying breaks. The retardation due to break is 2.5 m / s2

Answer 1

Answer:

25 m/s

Explanation:

Initial velocity, u=?  

Final velocity, v=0m/s (car is stopped)  

Acceleration, a=−2.5m/s  

2

 (negative because it is retardation)

Time, t=10s  

Use the relation of the first equation of motion:

v=u+at

0=u+(−2.5)×10

u=25m/s

Answer 2

Answer:

25 m/s.

Explanation:

Let the,

Initial velocity be u m/s.

Final velocity = v = 0 m/s [Since The car finally stops]

Acceleration = a = (-2.5) m/s^2

Time = t = 10 seconds.

ATQ,

v-u/t = a

=> 0-u/10 = -2.5

=>-u = (-2.5×10)

=>-u = -25.0

=>u = 25 m/s.[Ans]