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 $15s$ 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

Answer: (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
$s_{car}=ut+\frac{1}{2}a{t}^2$
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 $t=15s$ and acceleration $=\frac{\Delta v}{\Delta t}=\frac{45}{15}=\frac{9}{3}=3m/s^2$
Therefore, $s_{\text{car }}=\frac{1}{2}\times 3\times (15{)}^2=337.5m$
Now displacement of scooter is given as $s_{\text{scooner }}=v\times t$
From graph, we get $v=30m/s$, therefore, $s_{\text{sceoter }}=30\times 15=450m$
Thus, the difference between the distance travelled by car and scooter in 15 seconds is
$\Delta s=s_{\text{seooter }}-s_{\text{car }}=450-337.5=112.5m$
$\Rightarrow \Delta s=112.5m$
(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 $v_s=30m/s$
Therefore, displacement of scooter in time $t=30t$...(3)
Now, the car has uniform accelerated motion till time $t=15s$ 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 $u=0m/s,t_1=15s$ and $a=3m/s^2\Rightarrow v_c=45m/s$
Therefore, displacement of car in time $t=a+\frac{1}{2}\times 3(15{)}^2+45(t-15)$
$=337.5+45t-675$
Therefore, displacement of car in time $t=-337.5+45t$...(4)
Using relation (2) and equations (3) and (4), we get
$30t=-337.5+45t\Rightarrow (45-30)t=337.5$
$\Rightarrow t=\frac{337.5}{15}=22.5s$
$\Rightarrow t=22.5s$