A car is traveling along a straight horizontal road at a constant speed of 20\text{ m/s}. … — JAMB UTME

A car is traveling along a straight horizontal road at a constant speed of $20\text{ m/s}$ . The driver suddenly spots an obstacle ahead and takes $0.5\text{ s}$ to react before applying the brakes. Once the brakes are applied, the car decelerates uniformly at a rate of $4\text{ m/s}^2$ until coming to a complete stop. What is the total distance traveled by the car from the moment the driver spots the obstacle until the car stops completely?

A
$50\text{ m}$
$60\text{ m}$ Answer
C
$110\text{ m}$
D
$40\text{ m}$

Answer

The total distance traveled by the car is

60\text{ m}

The motion consists of two distinct stages: first, constant speed motion during the

0.5\text{ s}

reaction time giving

s_1 = 20 \times 0.5 = 10\text{ m}

; second, uniform deceleration from

20\text{ m/s}

to rest over

s_2 = \frac{u^2}{2a} = \frac{400}{8} = 50\text{ m}

. Adding both stages yields a total distance of

60\text{ m}

Step-by-Step Solution

Calculate the reaction distance traveled at constant speed before braking.

s_1 = v \times t = 20\text{ m/s} \times 0.5\text{ s} = 10\text{ m}

During the reaction time, acceleration is zero, so distance equals speed multiplied by time.

Calculate the braking distance using the third equation of motion.

Using

v^2 = u^2 + 2as

0 = (20)^2 + 2(-4)s_2 \implies 8s_2 = 400 \implies s_2 = 50\text{ m}

The car decelerates from

u = 20\text{ m/s}

v = 0\text{ m/s}

a = -4\text{ m/s}^2

Sum the reaction distance and the braking distance to obtain the total stopping distance.

s_{\text{total}} = s_1 + s_2 = 10\text{ m} + 50\text{ m} = 60\text{ m}

Total distance is the sum of distances covered in both stages of motion.

Key Concept

Multi-stage linear motion combining constant velocity reaction distance and uniform deceleration braking distance.

Estimated Time:1m 30s

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