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Question: A motorcycle and a car start from rest from the same place at the same time and travel in the same d...

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

A

20 s

B

25 s

C

30 s

D

35 s

Answer

35 s

Explanation

Solution

When car overtakes motorcycle, both have travelled the same distance in the same time. Let the total distance traveled be S and the total time taken to overtake be t. For motor cycle :

Maximum speed attained=36kmh−1= 36kmh^{- 1}

=36×51810ms−1= 36 \times \frac{5}{18}10ms^{- 1}

Since its acceleration =1.0ms−2,= 1.0ms^{- 2}, the time t1t_{1}taken by it to attain the maximum speed is given by

v=u+at1v = u + at_{1} (∵u=0\because u = 0)

10=0+1.0×t110 = 0 + 1.0 \times t_{1}

t1=10st_{1} = 10s

The distance covered by motorcycle in attaining the maximum speed is

S1=0+12at12=12×1.0×(10)2=50mS_{1} = 0 + \frac{1}{2}at_{1}^{2} = \frac{1}{2} \times 1.0 \times (10)^{2} = 50m

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

The distance covered by the motorcycle during this time is

S1=10×(t−10)=(10t−100)mS_{1} = 10 \times (t - 10) = (10t - 100)m

∴\thereforeTotal distance travelled by motorcycle in time t is

S=S1+S1=50+(10t−100)S = S_{1} + S_{1} = 50 + (10t - 100)

= (10t - 50)m …… (i)

For car

Maximum speed attained =54kmh−1= 54kmh^{- 1}

=54×518=15ms−1= 54 \times \frac{5}{18} = 15ms^{- 1}

Since its accelerations =0.5ms−2= 0.5ms^{- 2}

The times taken by it to attain the maximum speed is given by

15=0+0.5=t215 = 0 + 0.5 = t_{2} (∵u=0\because u = 0)

Or t2=30st_{2} = 30s

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

S2=0+12at23=12×0.5×(30)2=225mS_{2} = 0 + \frac{1}{2}at_{2}^{3} = \frac{1}{2} \times 0.5 \times (30)^{2} = 225m

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

The distance covered by the car during this time is

S2=15×(t−30)=(15t−450)mS_{2} = 15 \times (t - 30) = (15t - 450)m

∴\thereforeTotal distance travelled by car in time t is

S=S2+S2=225+(15t−450)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 = 175 or t = 35s