A traffic officer on a stationary motorcycle spots a car moving past at a constant speed o… — JAMB UTME

A traffic officer on a stationary motorcycle spots a car moving past at a constant speed of $30\text{ m/s}$ . The officer takes $2\text{ s}$ of reaction time before accelerating uniformly at $4\text{ m/s}^2$ to a maximum cruise speed of $40\text{ m/s}$ , after which the motorcycle continues at this constant speed. What is the total distance traveled by the motorcycle from its initial position to the point where it catches up with the car?

A
$360\text{ m}$
B
$510\text{ m}$
C
$600\text{ m}$
$840\text{ m}$ Answer

Answer

840\text{ m}

The correct answer is

840\text{ m}

. Over the first

2\text{ s}

, the motorcycle is stationary while the car covers

60\text{ m}

. Over the next

10\text{ s}

, the motorcycle accelerates to

40\text{ m/s}

, covering

200\text{ m}

, while the car travels another

300\text{ m}

(totaling

360\text{ m}

). To close the remaining

160\text{ m}

gap at a relative speed of

10\text{ m/s}

requires an additional

16\text{ s}

. The total elapsed time of

28\text{ s}

yields a total catch-up distance of

30\text{ m/s} \times 28\text{ s} = 840\text{ m}

Step-by-Step Solution

Calculate car position and motorcycle position at the end of the reaction time period (

t = 2\text{ s}

During the reaction time

t_r = 2\text{ s}

, the motorcycle remains stationary (

s_m = 0\text{ m}

). The car travels a distance

s_c = 30\text{ m/s} \times 2\text{ s} = 60\text{ m}

The motorcycle does not begin accelerating until after the officer's reaction time elapses.

Determine the time and distance required for the motorcycle to reach its maximum speed of

40\text{ m/s}

Time to reach maximum speed:

t_{acc} = \frac{v_{max} - u}{a} = \frac{40 - 0}{4} = 10\text{ s}

. Distance during acceleration:

s_{acc} = \frac{v_{max}^2 - u^2}{2a} = \frac{40^2 - 0}{2(4)} = 200\text{ m}

The motorcycle accelerates uniformly from rest until reaching its capped maximum velocity.

Calculate total positions at

t_{total1} = 2\text{ s} + 10\text{ s} = 12\text{ s}

from the instant the car passed.

Motorcycle position:

x_m(12) = 200\text{ m}

. Car position:

x_c(12) = 30\text{ m/s} \times 12\text{ s} = 360\text{ m}

. Distance gap remaining:

\Delta x = 360\text{ m} - 200\text{ m} = 160\text{ m}

Comparing positions at

12\text{ s}

establishes the remaining distance gap to be closed during the constant speed phase.

Calculate the time required during the constant-speed phase to close the remaining gap and find the total meeting distance.

Relative speed:

v_{rel} = 40\text{ m/s} - 30\text{ m/s} = 10\text{ m/s}

. Additional time needed:

\Delta t = \frac{160\text{ m}}{10\text{ m/s}} = 16\text{ s}

. Total elapsed time:

t = 12\text{ s} + 16\text{ s} = 28\text{ s}

. Catch-up distance:

s_{total} = 30\text{ m/s} \times 28\text{ s} = 840\text{ m}

With both vehicles moving at constant speeds, the relative velocity determines how quickly the remaining separation is eliminated.

Key Concept

Multi-stage linear motion involving delayed reaction time, uniform acceleration, and constant speed pursuit.

Estimated Time:3m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic