Question

A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of $72km/h$. The jeep follows it at a speed of $90km/h$, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike?
$A)\text{ }3km$
$B)\text{ 5}km$
$C)\text{ 1}km$
$D)\text{ 7}km$

Answer 1

Hint: We can find out the required distance by considering the relative velocity of the jeep with respect to the bike and find the time required for the jeep to catch up. Using this time, we can find out the distance travelled by the bike before getting caught, that is, the required distance from the turning.
Formula used:
${{v}_{12}}={{v}_{2}}-{{v}_{1}}$
$\text{Distance = Speed}\times \text{Time}$

Complete step-by-step answer:
We will find the relative velocity of the jeep with respect to the bike and therefore, find the time taken by the jeep to catch up. Using this time we will get the required distance.
Therefore, let us analyze the question.
The speed of the bike is ${{v}_{b}}=72km/h=72\times \dfrac{5}{18}=20m/s$ $\left( \because 1km/h=\dfrac{5}{18}m/s \right)$
The speed of the car is ${{v}_{c}}=90km/h=90\times \dfrac{5}{18}=25m/s$ $\left( \because 1km/h=\dfrac{5}{18}m/s \right)$
Let the time required by the jeep to catch up with the bike be $t$.
The initial time gap between the bike and the jeep is ${{t}_{0}}=10s$.
Let the initial distance gap between the jeep and the bike be ${{s}_{0}}$.
The initial distance is the distance moved by the bike from the turn in ${{t}_{0}}$.
Now,
$\text{Distance = Speed}\times \text{Time}$ --(1)
Therefore, using (1), we get
${{s}_{0}}={{v}_{b}}\times {{t}_{0}}=20\times 10=200m$ --(2)
Now, let the relative velocity of the jeep with respect to the bike be ${{v}_{bj}}$.
Now, the relative velocity ${{v}_{12}}$ of a body moving with velocity ${{v}_{2}}$ with respect to a body moving with velocity ${{v}_{1}}$ is given by
${{v}_{12}}={{v}_{2}}-{{v}_{1}}$ --(3)
Hence, using (3), we get
${{v}_{bj}}={{v}_{j}}-{{v}_{b}}=25-20=5m/s$ --(4)
Now, using the relative velocity, we can find out the time $t$ in which the jeep catches up with the bike, that is covers the initial distance gap between them.
Therefore, using (1), we get
${{s}_{0}}={{v}_{bj}}\times t$
$\therefore t=\dfrac{{{s}_{0}}}{{{v}_{bj}}}$
Using (2) and (4) in the above equation, we get
$t=\dfrac{200}{5}=40s$ --(5)
In time $t=40s$, the bike moves a further distance $s'$ from the turn.
Using (1), we get
$s'={{v}_{b}}\times t=20\times 40=800m$ [Using (5)] --(6)
Hence, the total distance $s$ from the turn will be the sum of the initial distance of the bike from the turn and the extra distance travelled by the bike before getting caught.
$\therefore s=s'+{{s}_{0}}$
Using (2) and (6), we get
$s=800+200=1000m=1km$ $\left( \because 1000m=1km \right)$
Therefore, the required distance from the turn after which the bike gets caught is $1km$.
Therefore, the correct option is $C)\text{ 1}km$.

Note: Students often forget that the total distance from the turn will be the sum of the initial distance of the bike from the turn and the extra distance travelled by it before getting caught. They only think that the distance from the turn will be the distance travelled by the bike before getting caught but they must remember that the bike had already travelled a distance ${{s}_{0}}$ at the start of the question which was implied by the fact that the jeep crossed the turn after $10s$ of the bike crossing the turn.