Question

A car starts from rest, attains 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 deceleration of 0.1 m s^-2. The total time of travel of the car is

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

Answer 1

Answer: (A)

1050 s

Answer 2

Let the car be accelerated from A to B. It move s with uniform velocity form B to C and then moves we uniform decelerations form C to D. For the motions of car from A to B.

$u = 0,v = 36kmh^{- 1} = 36 \times \frac{5}{18}ms^{- 1} = 10ms^{- 1},a = 0.2ms^{- 2}$

Time taken , $t_{1} = \frac{v - u}{a} = \frac{10ms^{- 1} - 0}{0.2ms^{- 2}} = 50s$

For the motion of car from B to C

S = 9 km = 9000 m

Time taken, $t_{2} = \frac{9000m}{10ms^{- 1}}$=900s

For the motions of car from C to D.

$v = 0,u = 10ms^{- 1},a = - 0.1ms^{- 2}$

Time taken, $t_{3} = \frac{0 - 10ms^{- 1}}{- 0.1ms^{- 2}} = 100s$

Total time taken$= t_{1} + t_{2} + t_{3}$

$50s + 900s + 100s = 1050s$