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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?
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The correct answer is:
$1.6 \mathrm{~s}$
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
$=\frac{s}{V_{A B}}=\frac{8}{5}=1.6 \mathrm{~s}$
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
$=\frac{s}{V_{A B}}=\frac{8}{5}=1.6 \mathrm{~s}$
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