A motorcycle and a car start from rest from the same place at the same time and travel in the same direction. The motorcycle accelerates at 1.0 m s - 2 up to a speed of 36 km h - 1 and the car at 0.5 m s - 2 up to a speed of 54 km h - 1 . The time

Question

A motorcycle and a car start from rest from the same place at the same time and travel in the same direction. The motorcycle accelerates at 1.0 m $s^{- 2}$ up to a speed of 36 km $h^{- 1}$ and the car at 0.5 m $s^{- 2}$ up to a speed of 54 km $h^{- 1}$ . The time at which the car would overtake the motorcycle is

Smart Achievers · Accepted Answer

(4)

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 motorcycle:

Maximum speed attained $= 36 km h^{- 1} = 36 \times \frac{5}{18} = 10 m s^{- 1}$

Since its acceleration is $1.0 m s^{- 2}$ , the time $t_{1}$ taken by it to attain the maximum speed is given by

$v = u + a t_{1} \Rightarrow 10 = 0 + 1.0 \times t_{1} \Rightarrow t_{1} = 10 s$ $(∵ u = 0)$

The distance covered by motorcycle in attaining the maximum speed is

$S_{1} = 0 + \frac{1}{2} a t_{1}^{2} = \frac{1}{2} \times 1.0 \times {(10)}^{2} = 50 m$

The time during which the motorcycle moves with maximum speed is $(t - 10) s$ .

The distance covered by the motorcycle during this time is $S_{1}' = 10 \times (t - 10) = (10 t - 100) m$

$∴$ Total distance travelled by motorcycle in time $t$ is

$S = S_{1} + S_{1}' = 50 + (10 t - 100) = (10 t - 50) m$ $\dots (i)$

For car:

Maximum speed attained $= 54 km h^{- 1} = 54 \times \frac{5}{18} = 15 m s^{- 1}$

Since its acceleration is $0.5 m s^{- 2}$ , the time taken by it to attain the maximum speed is given by

$15 = 0 + 0.5 \times t_{2}$ or $t_{2} = 30 s$ $(∵ u = 0)$

The distance covered by the car in attaining the maximum speed is

$S_{2} = 0 + \frac{1}{2} a t_{2}^{2} = \frac{1}{2} \times 0.5 \times {(30)}^{2} = 225 m$

The time during which the car moves with maximum speed is $(t - 30) s$

The distance covered by the car during this time is

$S_{2}' = 15 \times (t - 30) = (15 t - 450) m$

$∴$ Total distance travelled by car in time $t$ is

$S = S_{2} + S_{2}' = 225 + (15 t - 450) = (15 t - 225) m$ $\dots (i i)$

From equations (i) and (ii), we get

$10 t - 50 = 15 t - 225$ or $5 t = 175$ or $t = 35 s$

Solution

1 20 s

2 25 s

3 30 s

4 35 s

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