A police jeep is chasing with a velocity of $45 ext{ km/h}$ a thief in another jeep moving with a velocity o

Question

A police jeep is chasing with a velocity of $45 	ext{ km/h}$ a thief in another jeep moving with a velocity of $153 	ext{ km/h}$. Police fire a bullet with a muzzle velocity of $180 	ext{ m/s}$. The velocity with which it will strike the car of the thief is:

Accepted Answer

1.  **Convert Velocities to SI Units (m/s):**
    *   Velocity of Police Jeep ($v_p$) = $45 	ext{ km/h} = 45 	imes \frac{5}{18} = 12.5 	ext{ m/s}$.
    *   Velocity of Thief's Jeep ($v_t$) = $153 	ext{ km/h} = 153 	imes \frac{5}{18} = 42.5 	ext{ m/s}$.
2.  **Effective Velocity of Bullet:** Since the bullet is fired from the moving police jeep, its velocity relative to the ground ($v_{b,g}$) is the sum of the muzzle velocity and the jeep's velocity.
    *   $v_{b,g} = 	ext{Muzzle Velocity} + v_p = 180 + 12.5 = 192.5 	ext{ m/s}$.
3.  **Calculate Relative Velocity:** The velocity with which the bullet strikes the thief's car is the relative velocity of the bullet with respect to the thief ($v_{b,t}$).
    *   Since both are moving in the same direction: $v_{b,t} = v_{b,g} - v_t$ .
    *   $v_{b,t} = 192.5 - 42.5 = 150 	ext{ m/s}$.

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