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Chapter 4: Problem 57

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? a. \(1 \mathrm{~s}\) b. \(2 \mathrm{~s}\) c. \(4 \mathrm{~s}\) d. \(16 \mathrm{~s}\)

Short Answer

Expert verified
b. 2 seconds

Step by step solution

01

Understand the Problem

We have a body sliding down a smooth inclined plane, taking 4 seconds to reach the bottom from rest. We want to find the time taken to travel only one fourth of the total distance starting from rest.
02

Kinematic Equation for Motion on Incline

Since the body is sliding from rest, we can use the kinematic equation for uniformly accelerated motion: \( s = ut + \frac{1}{2}at^2 \), where \( s \) is the distance traveled, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time taken. Here, \( u = 0 \).
03

Calculate Total Distance and Acceleration

Let's say the total distance the body slides is \( S \). From the equation \( S = \frac{1}{2}a(4)^2 \), we can find \( \frac{1}{2}a = \frac{S}{16} \).
04

Apply to One Fourth Distance

For one fourth of the distance, \( s = \frac{S}{4} \). Using the kinematic equation \( \frac{S}{4} = \frac{1}{2} a t_1^2 \), we substitute \( \frac{1}{2} a = \frac{S}{16} \), giving us \( \frac{S}{4} = \frac{S}{16} t_1^2 \).
05

Solve for \( t_1 \)

From \( \frac{S}{4} = \frac{S}{16} t_1^2 \), we simplify to \( 4 = \frac{t_1^2}{4} \), leading to \( t_1^2 = 16 \) and \( t_1 = 4 \).
06

Conclude Solution

This solution implies that the time \( t_1 \) taken to cover one fourth the distance is 2 seconds, that is \( t_1 = 2 \).

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Key Concepts

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

Uniformly Accelerated Motion
When we talk about "uniformly accelerated motion," we're referring to a scenario where an object's acceleration is constant. This means that its speed increases at the same rate over time. If you think about a car accelerating smoothly without any bumps in speed, that's uniformly accelerated motion. Key Points to Remember:
  • Acceleration remains constant throughout the motion.
  • Initial velocity is often given as zero (like our exercise where the object starts from rest).
  • Common examples include cars moving on a straight road with constant acceleration and objects sliding down a smooth slope like our inclined plane example.
Using the right kinematic equations, you can predict how long it will take an object to reach a certain speed or cover a certain distance under uniform acceleration.
Inclined Plane
An inclined plane, like the one in our exercise, is a flat surface that's tilted at an angle. This angle is what allows gravity to pull an object downward along the plane, instead of letting it drop straight down. Why Inclined Planes are Interesting:
  • They help us study motion because they allow for controlled acceleration.
  • Less force is needed to move an object up a plane than lifting it vertically.
  • Ideal for demonstrating principles of physics like gravity and friction.
In the example, the smoothness of the plane means there's minimal friction, allowing the body to slide down solely due to gravitational pull. By understanding the motion along an incline, you see how angles and gravity work together to create movement.
Kinematic Equations
Kinematic equations are handy tools that help us understand object motion under uniform acceleration. These equations connect variables like distance, speed, time, and acceleration together in a meaningful way.Important Equations to Know:
  • The standard equation: \[ s = ut + \frac{1}{2}at^2 \]which calculates the distance (s) travelled with initial velocity (u), acceleration (a), and time (t).
  • Other equations can calculate final velocity or relate these variables to each other.
In the exercise case, using the kinematic equation, we found the time taken to travel a specified distance. By inserting values like initial speed and acceleration, we calculate the motion specifics on the inclined plane, guiding us to solve how long it takes the body to travel a portion of the distance.

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Most popular questions from this chapter

Two trains, one travelling at \(15 \mathrm{~ms}^{-1}\) and other at \(20 \mathrm{~ms}^{-1}\), are heading towards one another along a straight track. Both the drivers apply brakes simultaneously when they are \(500 \mathrm{~m}\) apart. If each train has a retardation of \(1 \mathrm{~ms}^{-2}\), the separation after they stop is a. \(192.5 \mathrm{~m}\) b. \(225.5 \mathrm{~m}\) c. \(187.5 \mathrm{~m}\) d. \(155.5 \mathrm{~m}\)

A person travels along a straight road for the first half time with a velocity \(v_{1}\) and the second half time with a velocity \(v_{21}\). Then the mean velocity \(\bar{v}\) is given by a. \(\bar{v}=\frac{v_{1}+v_{2}}{2}\) b. \(\frac{2}{\bar{v}}=\frac{1}{v_{1}}+\frac{1}{v_{2}}\) c. \(\bar{v}=\sqrt{v_{1} v_{2}}\) d. \(\bar{v}=\sqrt{\frac{v_{2}}{v_{1}}}\)

A wooden block is dropped from the top of a cliff \(100 \mathrm{~m}\) high and simultaneously a bullet of mass \(10 \mathrm{gm}\) is fired from the foot of the cliff upwards with a velocity of \(100 \mathrm{~m} / \mathrm{s}\). The bullet and wooden block will meet after a time a. \(10 \mathrm{~s}\) b. \(0.5 \mathrm{~s}\) c. \(1 \mathrm{~s}\) d. \(7 \mathrm{~s}\)

A parachutist drops first freely from an aeroplane for \(10 \mathrm{~s}\) and then parachute opens out. Now he descends with a net retardation of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\). If he bails out of the plane at a height of \(2495 \mathrm{~m}\) and \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), his velocity on reaching the ground will be a. \(5 \mathrm{~m} / \mathrm{s}\) b. \(10 \mathrm{~m} / \mathrm{s}\) c. \(15 \mathrm{~m} / \mathrm{s}\) d. \(20 \mathrm{~m} / \mathrm{s}\)

Two balls of different masses \(m_{\mathrm{a}}\) and \(m_{\mathrm{b}}\) are dropped from two different heights, viz., a and \(\mathrm{b}\). The ratio of times taken by the two to drop through these distances is a. a:b b. b:a c. \(\sqrt{a}: \sqrt{b}\) d. \(a^{2}: b^{2}\)
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