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 $1.0m{s}^{-2}$ 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 motorcycle is

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

Answer 1

Answer: (D)

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 $=36km{h}^{-1}=36\times \frac{5}{18}=10m{s}^{-1}$
Since its acceleration $=1.0m{s}^{-2}$, the time $t_1$ taken by it to attain the maximum speed is given by
$v=u+a{t}_1$
$\Rightarrow 10-0+1.0\times t_1$
$\Rightarrow t_1=10s\left(\because u=0\right)$
The distance covered by motorcycle in attaining the maximum speed is $S_1=0+\frac{1}{2}a{t}_1^2=\frac{1}{2}\times 1.0\times {\left(10\right)}^2=50m$
The time during which the motorcycle moves with maximum speed is $(t-10)s$.
The distance covered by the motorcycle during this time is $S_1^{\prime }=10\times (t-10)=(10t-100)m$
$\therefore$ Total distance travelled by motorcycle in time $t$ is
$S=S_1+S_1^{\prime }=50+\left(10t-100\right)$
$=\left(10t-50\right)m\dots \left(i\right)$
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 taken by it to attain the maximum speed is given by
$15=0+0.5\times t_2$ or $t_2=30s\left(\because u=0\right)$
The distance covered by the car in attaining the maximum speed is $S_2=0+\frac{1}{2}a{t}_2^2=\frac{1}{2}\times 0.5\times {\left(30\right)}^2=225m$
The time during which the car moves with maximum speed is $(t-30)s$.
The distance covered by the car during this time is
$S_2^{\prime }=15\times \left(t-30\right)=\left(15t-450\right)m$
$\therefore$ Total distance travelled by car in time $t$ is
$S=S_2+S_2^{\prime }=225+\left(15t-450\right)$
$=\left(15t-225\right)m\dots \left(ii\right)$
From equations $(i)$ and $(ii)$, we get
$10t-50=151-225$ or $51=175$ or $1=35s$