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

Option: 1
20 s

Option: 2
25 s

Option: 3
30 s

Option: 4
35 s

Answers (1)

best_answer

When the car overtakes the motorcycle, both have traveled the same distance at the same time. Let the total distance traveled be S and the total time taken to overtake be t.

For motorcycle:

Maximum speed attained

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

Since its acceleration $=1m{{s}^{-2}}$

the time $t_{1}$ taken by it to attain the maximum speed is given by,

$v=u+a{{t}_{1}}$

$10=0+1{{t}_{1}}$

${{t}_{1}}=10s$

The distance covered by motorcycle in attaining the maximum speed is,

${{S}_{1}}=0+\frac{1}{2}at_{1}^{2}$

${{S}_{1}}=\frac{1}{2}\times 1\times {{10}^{2}}$

${{S}_{1}}=50m$

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

The distance covered by the motorcycle during this time is,

$S_{1}^{'}=10\times (t-10)=(10t-100)m$

$\therefore$ Total distance traveled by the motor cycle in time t is,

$\\S={{S}_{1}}+S_{1}^{'} \\ S=50+(10t-50)m$

For car:

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

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

$\\v=u+a{{t}_{2}}\\ \\15=0+0.5\times {{t}_{2}}\\$

${{t}_{2}}=30s$

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

${{S}_{2}}=0+\frac{1}{2}at_{2}^{2}$

${{S}_{1}}=\frac{1}{2}\times 0.5\times {{30}^{2}}$

${{S}_{2}}=225m$

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

The distance covered by the car during this time is,

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

Thus finally equating the two equation of S we get,

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

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