The velocity-time graphs of a car and a scoot | vedclass

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(C) From the graph:For the car,the velocity increases uniformly from $0$ to $45\,m/s$ in $15\,s$. The distance travelled by the car in $15\,s$ is the

Vedclass · Accepted Answer

$112.5\,m$ and $22.5\,s$

Vedclass · Answer

(C) From the graph:For the car,the velocity increases uniformly from $0$ to $45\,m/s$ in $15\,s$. The distance travelled by the car in $15\,s$ is the area under the triangle $OAC$,which is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 15\,s 	imes 45\,m/s = 337.5\,m$.For the scooter,the velocity is constant at $30\,m/s$. The distance travelled by the scooter in $15\,s$ is $30\,m/s 	imes 15\,s = 450\,m$.$(i)$ The difference in distance travelled in $15\,s$ is $450\,m - 337.5\,m = 112.5\,m$.$(ii)$ After $15\,s$,the car moves with a constant velocity of $45\,m/s$. Let the car catch up with the scooter at time $t$ (where $t > 15\,s$).Distance of scooter at time $t = 30t$.Distance of car at time $t = 337.5 + 45(t - 15)$.Equating the distances: $30t = 337.5 + 45t - 675$.$15t = 337.5 \Rightarrow t = 22.5\,s$.

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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