Kinematics

JEE Main level, single correct, medium

107 of 308

Question 107 of 308

The velocity-time graphs of a car and a scooter are shown in the figure. (i) the difference between the distance travelled by the car and the scooter in 15 s15 \mathrm{~s} and (ii) the time at which the car will catch up with the scooter are, respectively

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Answer: (C) 112.5 m112.5 \mathrm{~m} and 22.5 s22.5 \mathrm{~s}

Using equation, a=v−uta=\frac{v-u}{t} and
S=ut+12at2S=u t+\frac{1}{2} a t^2
Distance travelled by car in 15sec15 \mathrm{sec}
=12(45)15(15)2=\frac{1}{2} \frac{(45)}{15}(15)^2
=6752 m=\frac{675}{2} \mathrm{~m}
Distance traveled by scooter in 15 seconds =30×15=450(∵=30 \times 15=450(\because distance == speed ×\times time )) Difference between distance travelled by car and scooter in 15sec,450−337.5=112.5 m15 \mathrm{sec}, 450-337.5=112.5 \mathrm{~m} Let car catches scooter in time t;
6752+45(t−15)=30t\frac{675}{2}+45(t-15)=30 t
337.5+45t−675=30t337.5+45 \mathrm{t}-675=30 \mathrm{t}
⇒15t=337.5\Rightarrow \quad 15 \mathrm{t}=337.5
⇒t=22.5sec\Rightarrow \quad \mathrm{t}=22.5 \mathrm{sec}

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