Q.A police van moving on a highway with a speed of $30\ \text{km h}^{-1}$ fires a bullet at a thief's car speeding away in the same direction with a speed of $192\ \text{km h}^{-1}$. If the muzzle speed of the bullet is $150\ \text{m s}^{-1}$, with what speed does the bullet hit the thief's car? (Note: Obtain that speed which is relevant for damaging the thief's car).

Concept: Relative Velocity — the speed relevant for damage is the bullet’s speed relative to the thief’s car.

1. Convert all speeds to consistent units (m/s):

   Police van: $30\ \text{km/h} = 30 \times \frac{5}{18} = \frac{25}{3}\ \text{m/s}$

   Thief’s car: $192\ \text{km/h} = 192 \times \frac{5}{18} = \frac{160}{3}\ \text{m/s}$

   Muzzle speed of bullet relative to van: $150\ \text{m/s}$.

2. Bullet’s velocity relative to ground = van’s velocity + muzzle speed (same direction):

   $v_{b,g} = \frac{25}{3} + 150 = \frac{25 + 450}{3} = \frac{475}{3}\ \text{m/s}$. …

The key idea here is relative velocity. When two objects move in the same direction, the speed of one relative to the other is the difference of their speeds. The bullet is fired from a moving police van, so its speed relative to the ground is the van's speed plus the muzzle speed. The thief's car is also moving. The damage to the car depends on how fast the bullet approaches it — that is, the bullet's speed relative to the car.

Let’s work through it step by step.

1. Convert all speeds to consistent units.

   The muzzle speed is given in $\text{m s}^{-1}$, but the van and car speeds are in $\text{km h}^{-1}$. We need everything in the same unit — let’s use $\text{m s}^{-1}$.

   Conversion factor: $1\ \text{km h}^{-1} = \frac{1000}{3600}\ \text{m s}^{-1} = \frac{5}{18}\ \text{m s}^{-1}$.

   • Van speed: $30\ \text{km h}^{-1} = 30 \times \frac{5}{18} = \frac{150}{18} = \frac{25}{3}\ \text{m s}^{-1} \approx 8.33\ \text{m s}^{-1}$.
   • Thief's car speed: $192\ \text{km h}^{-1} = 192 \times \frac{5}{18} = \frac{960}{18} = \frac{160}{3}\ \text{m s}^{-1} \approx 53.33\ \text{m s}^{-1}$.

2. Find the bullet's speed relative to the ground.

   The bullet is fired forward from the moving van. Its speed relative to the ground is the van's speed plus the muzzle speed (since both are in the same direction).

3. Determine the bullet's speed relative to the thief's car. The car is moving in the same direction as the bullet. The relative speed of the bullet with respect to the car is the difference between their ground speeds: …

Concept: Composing Relative Velocities Directly

Method: Direct Relative-Velocity Composition (never compute the bullet's ground speed at all)

The standard route first finds the bullet's speed relative to the ground, then subtracts the thief's car's ground speed. This method instead composes relative velocities directly — van→car and bullet→van — using the velocity-addition rule $\vec v_{A,C} = \vec v_{A,B} + \vec v_{B,C}$, which skips the ground-frame intermediate step entirely.

Steps

1. Convert the two vehicle speeds to $\text{m s}^{-1}$ (the muzzle speed is already in SI units, and is already a relative-velocity quantity — see Step 2):

2. Recognise what "muzzle speed" already means. A muzzle speed is, by its very definition, the bullet's speed relative to the gun (i.e. relative to the van it's mounted on) — so $v_{b,\text{van}} = 150\ \text{m s}^{-1}$ is already a relative velocity; there is no need to first convert it to a ground-frame value.

3. Find the van's velocity relative to the car (both moving the same direction, so this is a plain subtraction — but notice it's van minus car, not bullet minus car):

   $v_{\text{van},\text{car}} = v_{\text{van}} - v_{\text{car}} = \frac{25}{3} - \frac{160}{3} = -\frac{135}{3} = -45\ \text{m s}^{-1}$ …