The velocity-time graphs of a car and a scooter are shown in the figure. (i) the difference between the…
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 $15 \mathrm{~s}$ and (ii) the time at which the car will catch up with the scooter are, respectively
$337.5 \mathrm{~m}$ and $25 \mathrm{~s}$
$225.5 \mathrm{~m}$ and $10 \mathrm{~s}$
$112.5 \mathrm{~m}$ and $22.5 \mathrm{~s}$
$112.5 \mathrm{~m}$ and $15 \mathrm{~s}$
Solution
Using equation, $a=\frac{v-u}{t}$ and
$
S=u t+\frac{1}{2} a t^2
$
Distance travelled by car in $15 \mathrm{sec}$
$=\frac{1}{2} \frac{(45)}{15}(15)^2$
$=\frac{675}{2} \mathrm{~m}$
Distance traveled by scooter in 15 seconds $=30 \times 15=450(\because$ distance $=$ speed $\times$ time $)$ Difference between distance travelled by car and scooter in $15 \mathrm{sec}, 450-337.5=112.5 \mathrm{~m}$ Let car catches scooter in time t;
$\frac{675}{2}+45(t-15)=30 t$
$337.5+45 \mathrm{t}-675=30 \mathrm{t}$
$\Rightarrow \quad 15 \mathrm{t}=337.5$
$\Rightarrow \quad \mathrm{t}=22.5 \mathrm{sec}$
