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Chapter 3: Problem 5

\(\Lambda\) particle has an initial velocity of \(9 \mathrm{~m} / \mathrm{s}\) due east and a constant acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\) due west. 'The distance covered by the particle in the Sth sec. of its motion is (a) Lero (b) \(0.5 \mathrm{~m}\) (c) \(2 \mathrm{~m}\) (d) None

Short Answer

Expert verified
The correct answer is (a) Lero, with displacement \(0\text{ m}\).

Step by step solution

01

Understand the problem

We have a \(\Lambda\) particle moving with an initial velocity \(u = 9 \text{ m/s}\) towards the east. It undergoes a constant acceleration \(a = -2 \text{ m/s}^2\) towards the west. We aim to determine the displacement during the 5th second of its motion.
02

Use the displacement formula for nth second

For motion with constant acceleration, the displacement in the nth second is given by the formula:\[ s_n = u + \frac{1}{2}a(2n-1) \]where \(s_n\) is the displacement in the nth second, \(u\) is the initial velocity, \(a\) is the acceleration, and \(n\) is the specific second we're interested in. Here, \(n = 5\).
03

Substitute the values into the formula

The initial velocity \(u = 9 \text{ m/s}\) and acceleration \(a = -2 \text{ m/s}^2\) are given. Substitute these and \(n = 5\) into the formula:\[s_5 = 9 + \frac{1}{2}(-2)(2 \times 5 - 1)\]Simplifying the expression inside gives us:\[s_5 = 9 + \frac{1}{2}(-2)(10 - 1)\]
04

Simplify the expression

Simplify the inside of the equation:\[s_5 = 9 + \frac{1}{2}(-2)(9)\]Next, compute the multiplication:\[s_5 = 9 + (-9)\]Now add the numbers:\[s_5 = 0\]
05

Draw the conclusion

The displacement of the \(\Lambda\) particle during the 5th second is \(0\text{ m}\). Hence, the correct multiple choice answer is option (a) Lero.

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Key Concepts

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

Constant Acceleration
Constant acceleration means that the rate at which an object's speed changes stays the same over time. It is not affected by time, and no matter how long the object moves, the acceleration value remains unchanged. This is why it is called "constant".
In this particular exercise, we have a \( \Lambda \) particle experiencing a constant acceleration of \(-2 \mathrm{~m/s^2}\) towards the west.
  • The acceleration's direction is opposite to that of the initial velocity.
  • Despite this consistent change in velocity, the acceleration remains at \(-2 \mathrm{~m/s^2}\).
  • This influences how the object's speed decreases each second since the acceleration is negative, slowing the particle over time.
Initial Velocity
Initial velocity refers to the speed and direction an object has when it starts moving. It is a crucial starting point for understanding motion since it shows where the object begins its path.
In the example of the \(\Lambda\) particle:
  • The initial velocity is \(9 \mathrm{~m/s}\) east.
  • This starting speed interacts with the constant acceleration to determine future positions of the particle.
  • The challenge lies in calculating the net effect of this initial velocity combined with the constant, opposite acceleration over time.
Without identifying the initial velocity, understanding how the motion evolves, especially when considering an opposite acceleration, would be difficult.
Equations of Motion
Equations of motion are mathematical tools used to describe an object's motion with respect to time. They involve factors like initial velocity, acceleration, and time to predict future positions and velocities.
In scenarios involving constant acceleration, these equations are especially useful:
  • They help determine the displacement, velocity, and other relevant quantities in the particle's motion.
  • In the step-by-step solution presented, the formula for displacement in the nth second is used.\[ s_n = u + \frac{1}{2}a(2n-1) \]
  • The equation centers on calculating how far the particle travels in a specific second, which is essential for exercises like this one.
By substituting given values into the formulas, one can extrapolate information about motion, such as the displacement during a specified time, confirming the importance of mastering the equations of motion.

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Most popular questions from this chapter

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The total acceleration of a particle in circular motion is \(5 \mathrm{~m} / \mathrm{s}^{2}\). If the speed of the particle decreases at a rate of \(3 \mathrm{~m} / \mathrm{s}^{2}\), assuming \(r=2 \mathrm{~m}\), the angular speed \(\omega\) of the particle relative to the centre of the circle is (a) \(1 \mathrm{rad} / \mathrm{s}\) (b) \(\sqrt{2} \mathrm{rad} / \mathrm{s}\) (c) \(2 \mathrm{rad} / \mathrm{s}\) (d) \(2 \sqrt{2} \mathrm{rad} / \mathrm{s}\)

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Mass \(A\) is released from the top of a frictionless inclined plane \(18 \mathrm{~m}\) long and reaches the bottom 3 s later: \(\Lambda\) t the instant when \(A\) is released, a second mass \(B\) is projected upwards along the plane from the bottom with a certain initial velocity. Mass \(B\) travels a distance up the plane, stops and returns to the bottom so that it arrives simultaneously with \(A\). The two masses do not collide. Initial velocity of \(A\) is (a) \(4 \mathrm{~ms}^{1}\) (b) \(5 \mathrm{~ms}^{1}\) (c) \(6 \mathrm{~ms}\) (d) \(7 \mathrm{~ms}^{1}\)
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