A traffic officer on a stationary motorcycle spots a car passing at a constant speed of 15… — JAMB UTME

A traffic officer on a stationary motorcycle spots a car passing at a constant speed of $15\text{ m/s}$ . The officer immediately pursues the car, accelerating uniformly from rest at $4\text{ m/s}^2$ for $5\text{ s}$ , after which the motorcycle continues at the constant speed attained. How long after setting off does the motorcycle overtake the car?

$10\text{ s}$ Answer
B
$7.5\text{ s}$
C
$5\text{ s}$
D
$15\text{ s}$

Answer

The total time required for the motorcycle to overtake the car is

10\text{ s}

During the first

5\text{ s}

, the motorcycle accelerates from rest at

4\text{ m/s}^2

20\text{ m/s}

, covering

50\text{ m}

. In that same interval, the car moving at

15\text{ m/s}

covers

75\text{ m}

, creating a

25\text{ m}

separation gap. Beyond

5\text{ s}

, the motorcycle maintains

20\text{ m/s}

and closes in on the car (

15\text{ m/s}

) at a relative speed of

5\text{ m/s}

. It takes

5\text{ s}

to close the

25\text{ m}

gap, making the total pursuit time

10\text{ s}

Step-by-Step Solution

Calculate the state of both vehicles at the end of the motorcycle's acceleration phase (

t_1 = 5\text{ s}

Motorcycle speed

v = u + at = 0 + 4(5) = 20\text{ m/s}

. Motorcycle distance

s_1 = \frac{1}{2} a t^2 = \frac{1}{2}(4)(5^2) = 50\text{ m}

. Car distance

s_{\text{car}} = v_{\text{car}} \times t = 15 \times 5 = 75\text{ m}

Determine the position gap and relative speed after the acceleration phase ends.

Determine the remaining distance gap and relative speed between the vehicles.

Separation gap

d = 75\text{ m} - 50\text{ m} = 25\text{ m}

. Relative speed

v_{\text{rel}} = 20\text{ m/s} - 15\text{ m/s} = 5\text{ m/s}

Both vehicles now move at constant velocities, so relative speed determines how quickly the gap closes.

Calculate the time taken in the second phase and sum for total time.

Phase 2 time

t_2 = \frac{d}{v_{\text{rel}}} = \frac{25}{5} = 5\text{ s}

. Total time

t = t_1 + t_2 = 5\text{ s} + 5\text{ s} = 10\text{ s}

The total duration of pursuit is the acceleration duration plus the constant velocity catch-up duration.

Key Concept

Multi-stage linear motion involving uniform acceleration followed by constant velocity.

Estimated Time:1m 30s

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