A motorcycle and a car start from rest from the same place at the same and travel in the same direction. The motorcycle acceleration at `1.0 m s^(-1)` up to a speed of `36 km h^(-1)` and the car at `0.5 m s^(-1)` up to a speed of `54 km h^(-1)`. The time at which the car would overtake the motorcycle is

Options

A. 20 s
B. 25 s
C. 30 s
D. 35 s

Answer

D

Explanation

When car overtakes motorcycle, both have travelled the same distance in the same time. Let the total distance travelled be S and the total time taken to overtake be t.
For motor cycle :
Maximum speed attained = `36 km h^(-1) = 36 xx 5/18 = 10 m s^(-1)`
Since its acceleration = `1.0 m s^(-2)`, the time `t_1` taken by it to attain the maximum speed is given by
`v = u + at_1 implies 10 = 0 + 1.0 xx t_1 implies t_1 = 10 s` (since u = 0)
The distance covered by motocycle in attaining the maximum speed is `S_1 = 0 + 1/2 at^(2)_1 = 1/2 xx 1.0 xx (10)^(2)` = 50 m
The time during which the motorcycle moves with maximum speed is (t - 10) s.
The distance covered by the motorcycle during the time is `S^(')_1 = 10 xx (t - 10) = (10t - 100)` m
`therefore` Total distance travelled by motorcycle in time t is
`S = S_1 + S^(')_1 = 50 + (10t - 100) = (10t - 50) m` ...(i)
For car :
Maximum speed attained = `54 km h^(-1) = 54 xx 5/18 = 15 m s^(-1)`
Since its acceleration = `0.5 m s^(-2)`
The time taken by it to attain the maximum speed is given by
15 = `0 + 0.5 xx t_2` or `t_2 = 30 s` (since u = 0)
The distance covered by the car in attaining the maximum speed is `S_2 = 0 + 1/2 at^(2)_2 = 1/2 xx 0.5 xx (30)^(2)` = 225 m
The time during which the car moves with maximum speed is (t - 30) s.
The distance covered by the car during this time is
`S^(')_2 = 15 xx (t - 30) = (15t - 450)` m
`therefore` Totoal distance travelled by car in time t is
`S = S_2 + S^(')_2 = 225 + (15t - 450) = (15t - 225)`m ...(ii)
From equations (i) and (ii), we get
`10t - 50 = 15t - 225` or `5t = 75` or t = 35 s