A car initially traveling at a constant speed of 15\text{ m/s} accelerates uniformly at 2\… — JAMB UTME

A car initially traveling at a constant speed of $15\text{ m/s}$ accelerates uniformly at $2\text{ m/s}^2$ for $5\text{ s}$ . It then maintains the acquired maximum speed for $10\text{ s}$ before coming to rest under uniform retardation in $4\text{ s}$ . What is the total distance covered by the car during the entire motion?

Answer: 400 m

Answer

The total distance covered by the car during the entire motion is

400\text{ m}

The total distance is calculated by summing the distances covered in the three distinct phases of motion: acceleration (

100\text{ m}

), uniform velocity (

250\text{ m}

), and uniform retardation (

50\text{ m}

), yielding a total distance of

400\text{ m}

Step-by-Step Solution

Calculate final speed and distance for the acceleration phase.

Final speed

v = 25\text{ m/s}

and distance

s_1 = 100\text{ m}

Using kinematic equations

v = u + a t_1 = 15 + (2)(5) = 25\text{ m/s}

and

s_1 = u t_1 + \frac{1}{2}a t_1^2 = 15(5) + \frac{1}{2}(2)(5^2) = 75 + 25 = 100\text{ m}

Calculate the distance covered during the constant speed phase.

Distance

s_2 = 250\text{ m}

The car maintains the acquired speed of

25\text{ m/s}

for

10\text{ s}

, giving

s_2 = v \cdot t_2 = 25 \times 10 = 250\text{ m}

Calculate the distance covered during the retardation phase.

Distance

s_3 = 50\text{ m}

Using average velocity for uniform retardation to rest:

s_3 = \frac{v + 0}{2} t_3 = \frac{25}{2} \times 4 = 50\text{ m}

Sum the distances from all three stages to determine total distance.

Total distance

S = 400\text{ m}

S = s_1 + s_2 + s_3 = 100 + 250 + 50 = 400\text{ m}

Key Concept

Multi-stage linear motion and equations of uniform acceleration

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