Question

A police van moving on a highway with a speed of $30 km/h$ fires a bullet at a thief's car speeding away in the same direction with a speed of $192 km/h$. If the muzzle speed of the bullet is $150 m/s$, with what speed does the bullet hit the thief's car?
A) \[95m{s^{ - 1}}\]
B) \[105 m{s^{ - 1}}\]
C) \[115 m{s^{ - 1}}\]
D) \[125 m{s^{ - 1}}\]

Answer 1

Hint: The relative velocities between two moving objects can find out by adding their respective velocities with respect to ground. If two objects are moving in opposite directions, the magnitude of the relative velocity of one object with respect to the other is equal to the sum of the magnitude of their velocities.

Complete step by step solution:
It is given that a police car is chasing a thief’s car. In the process a bullet is fired from the police car towards the thief car. Here we need to find out how much velocity the bullet hits the thief’s car.
Here we can see that the velocities of bullet, police car, and thief’s cars are given but their units are different. For solving any questions all the units need to be in the same system
So, we will convert all velocities into \[m{s^{ - 1}}\]
Now the velocity of the police car is ${V_p} = 30km{h^{ - 1}}$
We will convert it in \[m{s^{ - 1}}\]
So ${V_p} = 30 \times \dfrac{5}{{18}}$
$ \Rightarrow {V_p} = \dfrac{{25}}{3}m{s^{ - 1}}$
Now the velocity of the thief’s car is given as ${V_T} = 192km{h^{ - 1}}$
We will also convert it into \[m{s^{ - 1}}\]
So ${V_T} = 192 \times \dfrac{5}{{18}}$
$ \Rightarrow {V_T} = \dfrac{{160}}{3}m{s^{ - 1}}$
The muzzle velocity of the bullet is already in $m{s^{ - 1}}$so we do not need to change it
$\Rightarrow {V_B} = 150m{s^{ - 1}}$Now
The velocity of the bullet with respect to the police car will be the velocity of the bullet wrt respect to the ground minus velocity of the police car wrt to ground
$\Rightarrow {V_{Bp}} = {V_{BG}} - {V_{pG}}$
$\Rightarrow {V_{BG}} = {V_{Bp}} - {V_{pG}}$
$\Rightarrow {V_{BG}} = 150 + \dfrac{{25}}{3}$
$\Rightarrow {V_{BG}} = \dfrac{{475}}{3}m{s^{ - 1}}$
Now we need to find the velocity of the bullet with which it hits the thief’s car
So, we need to find the velocity of the bullet with respect to the thief
$\Rightarrow {V_{BT}} = {V_{BG}} - {V_{TG}}$
$\Rightarrow {V_{BT}} = \dfrac{{475}}{3} - \dfrac{{160}}{3}$
$\Rightarrow {V_{BT}} = 105m{s^{ - 1}}$

Final answer is (B), The bullet will hit the thief’s car with a velocity of $105m{s^{ - 1}}$.

Note: 1. All the units in the question need to be in the same system.
2. The direction of motion of all the three objects needs to be considered properly because velocity is a vector quantity and any reverse direction needs to be calculated with a negative sign.
3. The velocity of the object with respect to the ground is the real velocity of the object with which it is moving on the ground.