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

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

Answer 1

Answer: (D)

$35\,s$

Answer 2

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 $=36\,km\,h^{-1}=36\times\frac{5}{18}=10\,m\,s^{-1}$ Since its acceleration $= 1.0 \,m \,s^{-2}$, the time $t_1$ taken by it to attain the maximum speed is given by $v=u+at_{1}$ $\Rightarrow 10-0+1.0\times t_{1}$ $\Rightarrow t_{1}=10\,s\quad\left(\because u=0\right)$ The distance covered by motorcycle in attaining the maximum speed is $S_{1}=0+\frac{1}{2} at^{2}_{1}=\frac{1}{2}\times1.0\times\left(10\right)^{2}=50\,m$ The time during which the motorcycle moves with maximum speed is $(t - 10) \,s$. The distance covered by the motorcycle during this time is $S'_1 = 10 \times (t - 10) = (10t - 100)\, m$ $\therefore$ Total distance travelled by motorcycle in time $t$ is $S=S_{1}+S'_{1}=50+\left(10t-100\right)$ $=\left(10t-50\right)\,m\quad\ldots\left(i\right)$ For car : Maximum speed attained $=54\,km\,h^{-1}=54\times\frac{5}{18}=15\,m\,s^{-1}$ Since its acceleration $= 0.5\, m \,s^{-2}$ The time taken by it to attain the maximum speed is given by $15=0+0.5\times t_{2}$ or $t_{2}=30\,s\quad\left(\because u=0\right)$ The distance covered by the car in attaining the maximum speed is $S_{2}=0+\frac{1}{2}at^{2}_{2}=\frac{1}{2}\times0.5\times\left(30\right)^{2}=225\,m$ The time during which the car moves with maximum speed is $(t - 30) \,s$. The distance covered by the car during this time is $S'_{2}=15\times\left(t-30\right)=\left(15t-450\right)\,m$ $\therefore$ Total distance travelled by car in time $t$ is $S=S_{2}+S'_{2}=225+\left(15t-450\right)$ $=\left(15t-225\right)m\quad\ldots\left(ii\right)$ From equations $(i)$ and $(ii)$, we get $10t - 50 = 151 - 225$ or $51 = 175$ or $1 = 35 \,s$