An electric cart moving along a straight horizontal track at an initial speed of 10\text{ … — JAMB UTME

An electric cart moving along a straight horizontal track at an initial speed of $10\text{ m/s}$ accelerates uniformly at $3\text{ m/s}^2$ for a duration of $4\text{ s}$ . Immediately after this phase, it applies its brakes and decelerates uniformly at $2\text{ m/s}^2$ until it comes to a complete stop. What is the total distance, in meters, traveled by the cart during the entire motion?

Answer: 185 m

Answer

The total distance traveled by the cart during the entire motion is

185\text{ m}

The total distance is calculated by analyzing the two distinct stages of motion. In the first phase, traveling at an initial velocity of

10\text{ m/s}

with an acceleration of

3\text{ m/s}^2

for

4\text{ s}

yields a distance of

64\text{ m}

and a peak velocity of

22\text{ m/s}

. In the second phase, decelerating from

22\text{ m/s}

to rest at

2\text{ m/s}^2

requires a distance of

121\text{ m}

. Adding both distances (

64\text{ m} + 121\text{ m}

) yields the total distance of

185\text{ m}

Step-by-Step Solution

Determine the distance (

s_1

) and final velocity (

v_1

) during the uniform acceleration phase

s_1 = 64\text{ m}

and

v_1 = 22\text{ m/s}

Using kinematic formulas

s_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2

and

v_1 = u_1 + a_1 t_1

for constant acceleration.

Determine the stopping distance (

s_2

) during the uniform deceleration phase

s_2 = 121\text{ m}

Using

v_2^2 = v_1^2 - 2 a_2 s_2

with final speed

v_2 = 0\text{ m/s}

Sum the distances from both stages

S_{\text{total}} = 185\text{ m}

Total displacement for multi-stage motion along a straight line is the sum of displacements in each stage.

Key Concept

Multi-stage linear motion with constant acceleration and deceleration

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