A car starts from rest, attain a velocity of `36 km h^(-1)` with an acceleration of `0.2 m s^(-2)`, travels 9 km with this uniform velocity and then comes to halt with a uniform deaceleration of 0.1 `m s^(-2)`. The total time of travel of the car is

Options

A. 1050 s
B. 1000 s
C. 950 s
D. 900 s

Answer

A

Explanation

Let the car be accelerated from A to B. It moves with uniform velocity from B to C and then moves with uniform deacleration from C to D.
For the motion of car from A to B,
u = 0, v = `36 km h^(-1) = 36 xx 5/18 m s^(-1) = 10 m s^(-1) , a = 0.2 m s^(-2)`
Time taken, `t_1= v - u / a = 10 m s^(-1) - 0 / 0.2 m s^(-2)` = 50 s
For the motion of car from B to C
S = 9 km = 9000 m
Time taken, `t_3= 0 - 10 m s^(-1) / -0.1 m s^(-2)` = 100 s
Total time taken = `t_1 + t_2 + t_3, = 50 s + 900 s + 100 s = 1050` s