Question

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.

• A. 112.5 m and 22.5 s
• B. 337.5 m and 25 s
• C. 112.5 m and 15 s
• D. 225.5 m and 10 s

Answer 1

The Correct Option is A

From the given velocity-time graph we observe that the car is having uniformly accelerated motion while the scooter is having uniform motion, thus, displacement of car is given as $s_{ car }=u t+\frac{1}{2} a t^{2}$ where $u$ is the initial velocity and $t$ is the time and $a$ is acceleration. Now we know initial velocity is zero $\Rightarrow u=0 .$ Therefore, $s_{ cx }=\frac{1}{2} a t^{2}$...(1) Given $t=15 s$ and acceleration $=\frac{\Delta v}{\Delta t}=\frac{45}{15}=\frac{9}{3}=3\, m / s ^{2}$ Therefore, $s_{\text {car }}=\frac{1}{2} \times 3 \times(15)^{2}=337.5\, m$ Now displacement of scooter is given as $s_{\text {scooner }}=v \times t$ From graph, we get $v=30 m / s$, therefore, $s_{\text {sceoter }}=30 \times 15=450\, m$ Thus, the difference between the distance travelled by car and scooter in 15 seconds is $\Delta s=s_{\text {seooter }}-s_{\text {car }}=450-337.5=112.5\, m$ $\Rightarrow \Delta s=112.5\, m$ (ii) Let the car catch up with the scooter in time $t$ seconds, then Displacement of scooter in time $t=$ Displacement of car in time $t$...(2) Now, since the scooter has uniform motion, therefore, displacement of scooter $=v_{s} \times t$ We know $v_{s}=30 m / s$ Therefore, displacement of scooter in time $t=30\, t$...(3) Now, the car has uniform accelerated motion till time $t=15 s$ and uniform motion beyond that. Therefore, displacement of car in time $t=\left(u t_{1}+\frac{1}{2} a t_{1}^{2}\right)+v_{ c } t_{2}$ We know $u=0 m / s , t_{1}=15 s$ and $a=3 m / s ^{2} \Rightarrow v_{ c }=45\, m / s$ Therefore, displacement of car in time $t=a+\frac{1}{2} \times 3(15)^{2}+45(t-15)$ $=337.5+45 t-675$ Therefore, displacement of car in time $t=-337.5+45 t$...(4) Using relation (2) and equations (3) and (4), we get $30\, t=-337.5+45\, t \Rightarrow(45-30) t=337.5$ $\Rightarrow t=\frac{337.5}{15}=22.5\, s$ $\Rightarrow t=22.5\, s$