Question

A particle has an initial velocity of 9 m/s due east and a constant acceleration of 2 m/s² due west. The distance covered by the particle in the fifth second of its motion is

• A. zero
• B. 0.5 m
• C. 2 m
• D. none of these

Answer 1

Answer: (B)

0.5 m

Answer 2

To solve this problem, we need to determine the distance covered by the particle in the fifth second of its motion. The fifth second refers to the time interval from $t=4$ s to $t=5$ s.

Given: Initial velocity, $u = 9 \text{ m/s}$ due east. Constant acceleration, $a = 2 \text{ m/s}^2$ due west.

Let's define the east direction as positive. So, $u = +9 \text{ m/s}$. Since acceleration is due west, it is in the opposite direction to the initial velocity, meaning it's a retardation. Thus, $a = -2 \text{ m/s}^2$.

First, we need to check if the particle changes its direction of motion within the fifth second (i.e., between $t=4$ s and $t=5$ s). A particle changes direction when its velocity becomes zero. The velocity of the particle at time $t$ is given by $v = u + at$. Set $v = 0$ to find the time when the particle momentarily stops: $0 = 9 + (-2)t$ $2t = 9$ $t = 4.5 \text{ s}$

Since $t = 4.5 \text{ s}$ falls within the interval of the fifth second ($[4 \text{ s}, 5 \text{ s}]$), the particle stops and reverses its direction of motion at $t=4.5$ s. Therefore, the distance covered is not simply the magnitude of the displacement in the fifth second. We must calculate the distance covered in two parts:

1. From $t=4$ s to $t=4.5$ s (moving east).
2. From $t=4.5$ s to $t=5$ s (moving west).

Part 1: Distance covered from $t=4$ s to $t=4.5$ s First, find the velocity of the particle at $t=4$ s: $v_4 = u + a(4)$ $v_4 = 9 + (-2)(4)$ $v_4 = 9 - 8 = 1 \text{ m/s}$ (due east)

The time interval for this part is $\Delta t_1 = 4.5 \text{ s} - 4 \text{ s} = 0.5 \text{ s}$. The displacement during this interval is given by $S = v_4 \Delta t_1 + \frac{1}{2}a(\Delta t_1)^2$: $S_{4 \to 4.5} = 1(0.5) + \frac{1}{2}(-2)(0.5)^2$ $S_{4 \to 4.5} = 0.5 - 1(0.25)$ $S_{4 \to 4.5} = 0.5 - 0.25 = 0.25 \text{ m}$ (due east) Since the velocity is positive in this interval, the distance covered is $|S_{4 \to 4.5}| = 0.25 \text{ m}$.

Part 2: Distance covered from $t=4.5$ s to $t=5$ s At $t=4.5$ s, the particle's velocity is $0 \text{ m/s}$. The time interval for this part is $\Delta t_2 = 5 \text{ s} - 4.5 \text{ s} = 0.5 \text{ s}$. The displacement during this interval is given by $S = v_{4.5} \Delta t_2 + \frac{1}{2}a(\Delta t_2)^2$: $S_{4.5 \to 5} = 0(0.5) + \frac{1}{2}(-2)(0.5)^2$ $S_{4.5 \to 5} = 0 - 1(0.25)$ $S_{4.5 \to 5} = -0.25 \text{ m}$ (due west) The distance covered in this part is the magnitude of the displacement, which is $|S_{4.5 \to 5}| = 0.25 \text{ m}$.

Total distance covered in the fifth second The total distance covered in the fifth second is the sum of the distances covered in Part 1 and Part 2: Total distance = $0.25 \text{ m} + 0.25 \text{ m} = 0.5 \text{ m}$.