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Chapter 4: Problem 36

The velocity-time relation of an electron starting from rest is given by \(v=k t\) where \(k=2 \mathrm{~m} / \mathrm{s}^{2}\). The distance traversed in first \(3 \mathrm{~s}\) is a. \(9 \mathrm{~m}\) b. \(16 \mathrm{~m}\) c. \(27 \mathrm{~m}\) d. \(36 \mathrm{~m}\)

Short Answer

Expert verified
The distance traversed by the electron in the first 3 seconds is 9 meters (option a).

Step by step solution

01

Understand the Problem

We are given the velocity-time relation of an electron which starts from rest: \( v = kt \), where \( k = 2 \mathrm{~m/s^2} \). We need to find the distance traversed by the electron in the first 3 seconds.
02

Recall the Distance Formula

The distance \( s \) travelled by an object with velocity \( v(t) \) is the integral of the velocity with respect to time, which is \( s = \int v(t) \, dt \).
03

Substitute the Velocity Function

Substitute the given velocity function \( v = kt \) into the distance formula: \[ s = \int_0^3 kt \, dt \] Here, the limits are from 0 to 3 seconds.
04

Perform the Integration

Perform the integration: \[ s = \int_0^3 2t \, dt = 2 \int_0^3 t \, dt \] The integral of \( t \) is \( \frac{t^2}{2} \), so:\[ s = 2 \left[ \frac{t^2}{2} \right]_0^3 = \left[ t^2 \right]_0^3 \] = \(3^2 - 0^2\) = 9.
05

Calculate the Distance

Substitute the limits to find \( s \): \( s = 9 \). Thus, the distance traversed in the first 3 seconds is 9 meters.

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Key Concepts

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

Understanding the Velocity-Time Relation
The velocity-time relation is crucial in understanding motion, particularly in one dimension. In this problem, the relation is given by the equation
  • \(v = kt\)
  • where \(k = 2 \, \mathrm{m/s^2}\)

This equation essentially tells us that the velocity \(v\) of the electron increases linearly with time \(t\). Starting from rest means that initially, when \(t=0\), the velocity is also zero.
As time progresses, the electron鈥檚 velocity increases at a constant rate of \(2 \, \mathrm{m/s^2}\). This relation helps us predict how fast the electron will be moving at any given second. For instance, at 1 second, the velocity is \(2 \, \mathrm{m/s^2} \times 1 \, \mathrm{s} = 2 \, \mathrm{m/s}\); at 2 seconds, it is \(4 \, \mathrm{m/s}\); and at 3 seconds, \(6 \, \mathrm{m/s}\). Understanding these changes over time is key to analyzing its subsequent distance traveled.
Calculating Distance Using the Distance Formula
Once we understand the velocity-time relation, we can move on to calculating the distance traveled. The general formula to find distance \(s\) when velocity \(v\) varies with time is:
  • \(s = \int v(t) \, dt\)

This formula tells us that the distance is essentially the area under the velocity-time curve from the initial time to the time of interest (in this case, from 0 to 3 seconds).
Inserting our velocity equation \(v = kt = 2t\) into this formula gives:
  • \(s = \int_0^3 2t \, dt\)
This new expression involves calculus, specifically the operation of integration, to find the total distance covered.
Solving the Problem Using the Integration Method
Integration is a powerful mathematical technique that helps us calculate areas under curves, which corresponds to distance in this context.
For our function \(2t\), integrating it involves finding the antiderivative, which is:
  • \(\int 2t \, dt = 2 \times \frac{t^2}{2}\)
  • simplifying to \(t^2\)

Next, apply the limits from 0 to 3 seconds, which means evaluating \([t^2]_0^3\):
  • \(3^2 - 0^2 = 9 - 0 = 9\)

Thus, the distance traveled by the electron in the first 3 seconds is 9 meters. This approach, using integration, not only provides the exact distance but also demonstrates the practical application of calculus in physics and motion studies.

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Most popular questions from this chapter

A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point \(h \mathrm{~m}\) below the point of projection is twice of the velocity at a point \(h \mathrm{~m}\) above the point of projection. Find the maximum height reached by the ball above the top of the tower. a. \(2 h\) b. \(3 h\) c. \((5 / 3) h\) d. \((4 / 3) h\)

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. \(72 \mathrm{~km}\)

A point moves with uniform acceleration and \(v_{1}, v_{2}\) and \(v_{3}\) denote the average velocities in the three successive intervals of time \(t_{1}, t_{2}\) and \(t_{3} .\) Which of the following relations is correct? a. \(\left(v_{1}-v_{2}\right):\left(v_{2}-v_{3}\right)=\left(t_{1}-t_{2}\right):\left(t_{2}+t_{3}\right)\) b. \(\left(v_{1}-v_{2}\right):\left(v_{2}-v_{3}\right)=\left(t_{1}+t_{2}\right):\left(t_{2}+t_{3}\right)\) c. \(\left(v_{1}-v_{2}\right):\left(v_{2}-v_{3}\right)=\left(t_{1}-t_{2}\right):\left(t_{1}-t_{3}\right)\) d. \(\left(v_{1}-v_{2}\right):\left(v_{2}-v_{3}\right)=\left(t_{1}-t_{2}\right):\left(t_{2}-t_{3}\right)\)

If \(x\) denotes displacement in time \(t\) and \(x=a \cos t\), then the acceleration is a. \(a \cos t\) b. \(-a \cos t\) c. \(a \sin t\) d. \(-a \sin t\)

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. \(20 \mathrm{~m} / \mathrm{s}\)
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