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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Retardation
Imagine you're riding a bike downhill and then start going uphill or applying brakes. The energy from going fast begins to wane, causing you to slow down. This concept is similar to what uniform retardation is in physics. Retardation is simply negative acceleration where the speed of an object decreases over time. In this problem, a constant retardation of \(3 \,\mathrm{ms}^{-2}\) means that the particle is slowing down at the same rate throughout its motion.

It's crucial to remember that retardation is a form of acceleration, but since it's a decrease in speed, it takes on a negative value. In scenarios where objects slow down uniformly, calculations remain simpler because the retardation, or deceleration, remains constant. This helps us apply the kinematic equations more straightforwardly.

Next time you hit the brakes in a car and feel the gradual halt, think of it as negative acceleration, applying the same principles of uniform retardation.
Equations of Motion
The equations of motion allow us to predict future states of moving objects based on their current state and acceleration. For a uniformly accelerated motion, including uniform retardation, these equations are essential tools:

  • \(v = u + at\)
  • \(s = ut + \frac{1}{2}at^2\)
  • \(v^2 = u^2 + 2as\)

Here, \(u\) is the initial velocity, \(v\) is the final velocity, \(a\) is the acceleration (or retardation), \(t\) is the time, and \(s\) is the displacement. In this problem, we used the second equation to calculate the distance.
For a particle with initial velocity \(12 \,\mathrm{ms}^{-1}\) and a uniform retardation of \(3 \,\mathrm{ms}^{-2}\), this equation becomes a tool to find out how much ground the particle covers in a particular time frame. This equation accurately informs us of the position of a particle at any moment.
Distance Calculation
Understanding how to compute the distance an object moves under uniform retardation requires us to break down the motion into segments and use one of the simplest equations of motion. Here, we calculate how far a particle travels by tracking its distance at specific intervals.

Using the formula \(s = ut + \frac{1}{2}at^2\), we insert the known values for each set period. At \(t_3 = 3 \, \mathrm{seconds}\), you get a distance of \(22.5 \, \mathrm{m}\). At \(t_4 = 4 \, \mathrm{seconds}\), the distance increases to \(24 \, \mathrm{m}\). Subtracting these gives how much the particle moved during the fourth second, providing the answer: \(1.5 \, \mathrm{m}\).

This step-by-step breakdown not only shows the calculation process but also the beauty of physics in predicting real-world phenomena through simple math. By combining initial conditions with uniform retardation, we determine the exact travel distance at any given moment in time.

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Most popular questions from this chapter

4\. A man can row a boat with speed \(4 \mathrm{~km} / \mathrm{hr}\) in still water. If the velocity of water in river is \(3 \mathrm{~km} / \mathrm{hr}\). The time taken to reach just opposite is (river width = \(500 \mathrm{~m})\) (A) \(\frac{500}{\sqrt{7}} \mathrm{hr}\) (B) \(\frac{1}{2 \sqrt{7}} \mathrm{hr}\) (C) \(100 \mathrm{hr}\) (D) None

A car is moving on a circular path of radius \(100 \mathrm{~m}\). Its speed \(v\) is changing with time as \(v=2 t^{2}\), where \(v\) in \(\mathrm{ms}^{-1}\) and \(t\) in second. The acceleration of car at \(t=5 \mathrm{~s}\) is approximately (A) \(20 \mathrm{~ms}^{-1}\) (B) \(25 \mathrm{~ms}^{-1}\) (C) \(30 \mathrm{~ms}^{-1}\) (D) \(32 \mathrm{~ms}^{-1}\)

If the range of a gun which fires a shell with muzzle speed \(V\) is \(R\), then the angle of elevation of the gun is (A) \(\cos ^{-1}\left(\frac{V^{2}}{R g}\right)\) (B) \(\cos ^{-1}\left(\frac{g R}{V^{2}}\right)\) (C) \(\frac{1}{2}\left(\frac{V^{2}}{R g}\right)\) (D) \(\frac{1}{2} \sin ^{-1}\left(\frac{g R}{V^{2}}\right)\)

A train starts from station \(A\) with uniform acceleration \(a_{1}\) for some distance and then goes with uniform retardation \(a_{2}\) for some more distance to come to rest at station \(B\). The distance between \(A\) and \(B\) is \(4 \mathrm{~km}\) and the train takes 4 hours to complete this journey. If acceleration and retardation are in \(\mathrm{km} /\) hour \(^{2}\), then (A) \(\frac{a_{1}}{a_{2}}=4\) (B) \(\frac{1}{a_{1}}+\frac{1}{a_{2}}=2\) (C) \(a_{1} a_{2}=1\) (D) None

A swimmer wishes to cross a \(800 \mathrm{~m}\) wide river flowing at \(6 \mathrm{~km} / \mathrm{hr}\). His speed with respect to water is \(4 \mathrm{~km} / \mathrm{hr}\). He crosses the river in shortest possible time. He is drifted downstream on reaching the other bank by a distance of (A) \(800 \mathrm{~m}\) (B) \(1200 \mathrm{~m}\) (C) \(400 \sqrt{13} \mathrm{~m}\) (D) \(2000 \mathrm{~m}\)
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