QJEEM18Phf337: JEE Main 2018 Physics - Class 11 JEEM18Phf337 Solution: The velocity-time graphs of a car and a scooter are shown in…

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.