/*! 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 22 A body sliding on a smooth incli... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 22

A body sliding on a smooth inclined plane requires \(4 \mathrm{~s}\) to reach the bottom, starting from rest at the top. How much time does it take to cover one-fourth the distance starting from rest at the top? (1) \(1 \mathrm{~s}\) (2) \(2 \mathrm{~s}\) (3) \(4 \mathrm{~s}\) (4) \(16 \mathrm{~s}\)

Short Answer

Expert verified
The body takes \(2 \mathrm{~s}\) to cover one-fourth the distance.

Step by step solution

01

Understanding the motion

The body is starting from rest, moving down a smooth incline. Its motion is uniformly accelerated since there is no friction. We need to find the time taken to travel one-fourth of the total distance covered in 4 seconds.
02

Using the equations of motion

Since the body starts from rest, the initial velocity \( u = 0 \). We use the equation \( s = ut + \frac{1}{2}at^2 \), where \( s \) is the distance, \( a \) is the acceleration, and \( t \) is time. When the body reaches the bottom in 4 seconds, the equation for the whole distance \( D \) is \( D = \frac{1}{2}a(4)^2 \). Simplifying, \( D = 8a \).
03

Calculating one-fourth the distance

One-fourth of the distance \( D \) is \( \frac{D}{4} = 2a \). We need to find the time \( t' \) to travel this distance. Thus, \( \frac{D}{4} = \frac{1}{2}a (t')^2 \) becomes \( 2a = \frac{1}{2}a (t')^2 \).
04

Solving for \( t' \)

From the equation \( 2a = \frac{1}{2}a (t')^2 \), we can solve for \( t' \). Canceling \( a \) from both sides and rearranging gives \( 2 = \frac{1}{2} (t')^2 \). Multiplying both sides by 2 yields \( 4 = (t')^2 \). Taking the square root gives \( t' = 2 \).

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Key Concepts

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

Uniform Acceleration
When a body moves under uniform acceleration, its velocity changes at a constant rate. This is a common situation in physics, such as an object sliding down a smooth inclined plane where there is no friction.
In this exercise, we're observing a body starting from rest and accelerating uniformly down an incline. The key point here is the absence of friction, which ensures that the acceleration remains constant.
  • Constant Acceleration: The same increase in velocity occurs during each unit of time.
  • Starting From Rest: Initial velocity, denoted as \( u \), is zero, making calculations easier.
Understanding that the body experiences constant acceleration is essential when using equations of motion to predict how long it takes to travel a given distance.
Equations of Motion
The equations of motion are fundamental tools in physics for describing how objects move under uniform acceleration. For any object moving with a constant acceleration \( a \), three key equations help us understand its motion.
In our scenario, the body begins from rest, so the initial velocity \( u = 0 \). This simplifies the primary equation of motion:
  • Equation of Motion: \( s = ut + \frac{1}{2}at^2 \)
  • Simplified Version for Start from Rest: \( s = \frac{1}{2}at^2 \)
This equation allows us to link time, acceleration, and distance, making it possible to solve problems involving inclined plane motion and other scenarios of uniform acceleration.
Distance Calculation in Physics
Calculating the distance traveled in physics requires an understanding of the body's initial conditions, and the forces acting on it. In this context, when given the total time and distance for a smooth, frictionless inclined plane, one can easily compute portions of that distance.
Given that the body takes 4 seconds to travel the total distance \( D \) under constant acceleration, we use the relationship derived from the equation of motion.
  • Total Distance: \( D = \frac{1}{2}a (4)^2 = 8a \)
  • One-Fourth Distance: \( \frac{D}{4} = 2a \)
The task requires finding the time to travel one-fourth of the distance, which involves solving \( 2a = \frac{1}{2}a(t')^2 \). By computing appropriately, one finds the time, \( t' = 2 \) seconds, demonstrating practical application of motion equations in solving real-world physics problems.

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Most popular questions from this chapter

The body will speed up if (1) Velocity and acceleration are in the same direction. (2) Velocity and acceleration are in opposite directions. (3) Velocity and acceleration are in perpendicular direction. (4) Velocity and acceleration are acting at acute angle w.r.t. each other.

An object is moving in the \(x-y\) plane with the position as a function of time given by \(\vec{r}=x(t) \hat{i}+y(t) \hat{j}\). Point \(O\) is at \(x\) \(=0, y=0\). The object is definitely moving towards \(O\) when (1) \(v_{x}>0, v_{y}>0\) (2) \(v_{x}<0, v_{y}<0\) (3) \(x v_{x}+y v_{y}<0\) (4) \(x v_{x}+y v_{y}>0\)

A car accelerates from rest at a constant rate of \(2 \mathrm{~ms}^{-2}\) for some time. Then it retards at a constant rate of \(4 \mathrm{~ms}^{-2}\) and comes to rest. It remains in motion for \(6 \mathrm{~s}\). (1) Its maximum speed is \(8 \mathrm{~m} \mathrm{~s}^{-1}\) (2) Its maximum speed is \(6 \mathrm{~ms}^{-1}\) (3) It travelled a total distance of \(24 \mathrm{~m}\). (4) It travelled a total distance of \(18 \mathrm{~m}\).

A drunkard is walking along a straight road. He takes five steps forward and three steps backward and so on. Each step is \(1 \mathrm{~m}\) long and takes \(1 \mathrm{~s}\). There is a pit on the road \(11 \mathrm{~m}\) away from the starting point. The drunkard will fall into the pit after (1) \(29 \mathrm{~s}\) (2) \(21 \mathrm{~s}\) (3) \(37 \mathrm{~s}\) (4) \(31 \mathrm{~s}\)

A particle starts from the origin with a velocity of \(10 \mathrm{~ms}^{-1}\) and moves with a constant acceleration till the velocity increases to \(50 \mathrm{~ms}^{-1}\). At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point? (1) Zero (2) \(10 \mathrm{~ms}^{-1}\) (3) \(50 \mathrm{~ms}^{-1}\) (4) \(70 \mathrm{~ms}^{-1}\)
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