A car travelling at $15 \mathrm{~m} / \mathrm{s}$ overtake another car travelling at $10 \mathrm{~m} / \mathrm{s}$. Assuming, each car is $4 \mathrm{~m}$ long. What is the time taken during the overtake?

• Correct answer
  A

  $1.6 \mathrm{~s}$
• B

  $0.8 \mathrm{~s}$
• C

  $0.6 \mathrm{~s}$
• D

  $0.4 \mathrm{~s}$

Solution

The motion of two car $A$ and $B$ is shown in the figure. Relative velocity of car $A$ w.r.t. car $B$ is $v_{A B}=v_A-v_B=15-10=5 \mathrm{~m} / \mathrm{s}$
Distance to be covered for overtake is given as $s=4 m+4 m=8 m$ Time taken by cars to cross $=$ Distance $/$ Speed $=\frac{s}{V_{A B}}=\frac{8}{5}=1.6 \mathrm{~s}$

Hence, the correct answer is option: 'A'.