/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 The distance travelled by a part... [FREE SOLUTION] | 91Ó°ÊÓ

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The distance travelled by a particle in a straight line motion is directly proportional to \(t^{1 / 2}\), where \(t=\) time elapsed. What is the nature of motion? a. Increasing acceleration b. Decreasing acceleration c. Increasing retardation d. Decreasing retardation

Short Answer

Expert verified
The nature of motion is decreasing retardation (option d).

Step by step solution

01

Identify the relation

The distance, \( s \), travelled by the particle is directly proportional to \( t^{1/2} \). This can be written as \( s = k \cdot t^{1/2} \), where \( k \) is a constant of proportionality.
02

Differentiate the distance to find velocity

Velocity \( v \) is the derivative of distance with respect to time \( t \). So, we calculate it as follows:\[v = \frac{ds}{dt} = \frac{d}{dt}(k \cdot t^{1/2}) = \frac{k}{2} \cdot t^{-1/2}.\]
03

Differentiate velocity to find acceleration

Acceleration \( a \) is the derivative of velocity with respect to time \( t \). So, we calculate it as follows:\[a = \frac{dv}{dt} = \frac{d}{dt}(\frac{k}{2} \cdot t^{-1/2}) = -\frac{k}{4} \cdot t^{-3/2}.\]
04

Analyze the sign of acceleration

The acceleration \( a = -\frac{k}{4} \cdot t^{-3/2} \) is negative for all positive \( t \). This indicates that the acceleration is decreasing over time since the magnitude of acceleration decreases as \( t \) increases, leading to a motion that becomes slower.
05

Determine the nature of motion

Since the acceleration is negative and decreases in magnitude as time increases, the motion is characterized by decreasing retardation. This means the speed reduces at a slower rate over time.

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Key Concepts

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

Straight Line Motion
When discussing straight line motion, we're focusing on movement along a single path without any deviation. In the original problem, a particle was moving in such a manner, with its distance covered being directly linked to the square root of time. This form of motion implies linear progression, where any change in movement is predictable and smooth.
In many real-world scenarios, straight line motion is the simplest representation of movement, often used in initial physics lessons to introduce concepts like distance, velocity, and acceleration. While real-world movements may be more complex, straight line motion helps students grasp fundamental principles easily.
Proportional Relationships
In the exercise, we see distance traveled by the particle is directly proportional to the square root of time. This implies if time doubles, the distance does not double but changes according to the square root relationship.
These relationships simplify subject understanding since one variable consistently affects another. When you know their proportional linkage, finding unknown quantities becomes easier - as illustrated with the formula:
  • Distance formula: \( s = k \cdot t^{1/2} \)
Here, the constant \( k \) represents how efficiently time translates into distance for this particular situation.
Differentiation in Kinematics
Differentiation is a powerful tool in kinematics for uncovering properties like velocity and acceleration from distance functions. In our example, we differentiate the distance function \( s = k \cdot t^{1/2} \) to get the velocity \( v = \frac{ds}{dt} \).
By performing this mathematical operation, we examine how the distance changes over time—essentially capturing the speed of the particle's movement:
  • Velocity formula: \( v = \frac{k}{2} \cdot t^{-1/2} \)
Continuous differentiation then allows us to move from velocity to acceleration, giving important insights about the particle's changing speed and direction over time.
Acceleration Analysis
Acceleration, or the rate of change of velocity, provides a deeper insight into motion dynamics. After differentiating the velocity, we found the acceleration function:
  • Acceleration formula: \( a = -\frac{k}{4} \cdot t^{-3/2} \)
This negative outcome signals decreasing acceleration—meaning the particle is slowing down over time. Such insights help us understand not just the speed but how the changes in speed affect motion patterns.
Thus, in this particular scenario, the decreasing negative acceleration, or decreasing retardation, suggests that although the particle's slowdown lessens, it still continues to lose speed as time advances. Understanding this concept helps students better predict and visualize kinematic behaviors in a straightforward manner.

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Most popular questions from this chapter

\(B_{1}, B_{2}\) and \(B_{3}\) are three balloons ascending with velocities \(v, 2 v\) and \(3 v\), respectively. If a bomb is dropped from each when they are at the same height, then a. bomb from \(B_{1}\) reaches ground first b. bomb from \(B_{2}\) reaches ground first c. bomb from \(B_{3}\) reaches ground first d. they reach the ground simultaneously

A point moves in a straight line so that its displacement \(x \mathrm{~m}\) at time \(t \mathrm{~s}\) is given by \(x^{2}=1+t^{2}\). Its acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) at time \(t\) s is a. \(\frac{1}{x^{3}}\) b. \(\frac{-t}{x^{3}}\) c. \(\frac{1}{x}-\frac{t^{2}}{x^{3}}\) d. \(\frac{1}{x}-\frac{1}{x^{2}}\)

The magnitude of the displacement is equal to the distance covered in a given interval of time if the particle a. moves with constant acceleration along any path b. moves with constant speed c. moves in same direction with constant velocity or with variable velocity d. moves with constant velocity

Two cars \(A\) and \(B\) are travelling in the same direction with velocities \(V_{A}\) and \(V_{B}\left(V_{A}>V_{B}\right) .\) When car \(A\) is at a distance s behind \(\operatorname{car} B\), the driver of \(\operatorname{car} A\) applies the brakes producing a uniform retardation \(a\); there will be no collision when a. \(s<\frac{\left(V_{A}-V_{B}\right)^{2}}{2 a}\) b. \(s=\frac{\left(V_{A}-V_{B}\right)^{2}}{2 a}\) c. \(s \geq \frac{\left(V_{A}-V_{B}\right)^{2}}{2 a}\) d. \(s \leq \frac{\left(V_{A}-V_{B}\right)^{2}}{2 a}\)

Two trains each travelling with a speed of \(37.5 \mathrm{~km} / \mathrm{h}\) are approaching each other on the same straight track. A bird that can fly at \(60 \mathrm{~km} / \mathrm{h}\) flies off from one train when they are \(90 \mathrm{~km}\) apart and heads directly for the other train. On reaching the other train it flies back to the first and so on. Total distance covered by the bird is a. \(90 \mathrm{~km}\) b. \(54 \mathrm{~km}\) c. \(36 \mathrm{~km}\) d. \(72 \mathrm{~km}\)

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