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Chapter 2: Problem 6

The initial velocity of a particle moving along a straight line is \(12 \mathrm{~ms}^{-1}\) and its retardation is \(3 \mathrm{~ms}^{-2}\). The distance moved by the particle in the fourth second of its motion is (A) \(1.5 \mathrm{~m}\) (B) \(22.5 \mathrm{~m}\) (C) \(24 \mathrm{~m}\) (D) \(72 \mathrm{~m}\)

Short Answer

Expert verified
The distance moved by the particle in the fourth second of its motion is \(1.5 \mathrm{~m}\).

Step by step solution

01

Identify given information

We are given the following information: - Initial velocity (u) = 12 m/s - Retardation (a) = -3 m/s^2 (negative value since it's slowing down)
02

Find the time at the end of the third second

Let t_3 be the time at the end of the third second. Since the motion starts at t = 0, we have t_3 = 3 seconds.
03

Find the time at the end of the fourth second

Let t_4 be the time at the end of the fourth second. Since the motion starts at t = 0, we have t_4 = 4 seconds.
04

Find the distance covered at the end of the third second

We can use the equation of motion under uniform retardation to find the distance covered at the end of the third second (s_3): s_3 = ut_3 + (1/2)at_3^2 = 12(3) + (1/2)(-3)(3)^2 = 36 - 13.5 = 22.5 m
05

Find the distance covered at the end of the fourth second

We can use the same equation to find the distance covered at the end of the fourth second (s_4): s_4 = ut_4 + (1/2)at_4^2 = 12(4) + (1/2)(-3)(4)^2 = 48 - 24 = 24 m
06

Find the distance moved in the fourth second

Now, we can find the distance moved in the fourth second by subtracting the distance covered at the end of the third second (s_3) from the distance covered at the end of the fourth second (s_4): Distance moved in the fourth second = s_4 - s_3 = 24 m - 22.5 m = 1.5 m So, the answer is (A) 1.5 m.

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