JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 6)
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 $$15$$ $$s$$ and (ii) the time at which the car will catch up with the scooter are, respectively.
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$$112.5$$ $$m$$ and $$22.5$$ $$s$$
$$337.5$$ $$m$$ and $$25$$ $$s$$
$$112.5$$ $$m$$ and $$15$$ $$s$$
$$225.5$$ $$m$$ and $$10$$ $$s$$
Explanation
Till 15 sec car has accelerative motion and scooter has constant velocity in entire motion.
Total Distance travelled by the car in 15 sec, = $${1 \over 2}$$ $$ \times $$ $${{45} \over {15}}$$ $$ \times $$ (15)2 = $${{675} \over 2}$$ m
Distance travelled by scooter in 15 sec.
= V $$ \times $$ t = 30 $$ \times $$ 15 = 450 m
$$\therefore\,\,\,\,$$ Difference between distance travelled by the car and scooter in 15 sec,
= 450 $$-$$ $${{675} \over 2}$$ = 112.5 m
Now, assume car catches the scooter in time t,
$$\therefore\,\,\,\,$$ Car travelled in t sec = scooter travelled in t sec.
$$ \Rightarrow $$ $$\,\,\,\,$$ $${{675} \over 2}$$ + 45(t$$-$$ 15) = 30 $$ \times $$ t
$$ \Rightarrow $$ $$\,\,\,\,$$ 337.5 + 45t $$-$$ 675 = 30t
$$ \Rightarrow $$ $$\,\,\,\,$$ 15 t = 337.5
$$ \Rightarrow $$ $$\,\,\,\,$$ t = 22.5 sec.
Total Distance travelled by the car in 15 sec, = $${1 \over 2}$$ $$ \times $$ $${{45} \over {15}}$$ $$ \times $$ (15)2 = $${{675} \over 2}$$ m
Distance travelled by scooter in 15 sec.
= V $$ \times $$ t = 30 $$ \times $$ 15 = 450 m
$$\therefore\,\,\,\,$$ Difference between distance travelled by the car and scooter in 15 sec,
= 450 $$-$$ $${{675} \over 2}$$ = 112.5 m
Now, assume car catches the scooter in time t,
$$\therefore\,\,\,\,$$ Car travelled in t sec = scooter travelled in t sec.
$$ \Rightarrow $$ $$\,\,\,\,$$ $${{675} \over 2}$$ + 45(t$$-$$ 15) = 30 $$ \times $$ t
$$ \Rightarrow $$ $$\,\,\,\,$$ 337.5 + 45t $$-$$ 675 = 30t
$$ \Rightarrow $$ $$\,\,\,\,$$ 15 t = 337.5
$$ \Rightarrow $$ $$\,\,\,\,$$ t = 22.5 sec.
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