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Chapter 2: Problem 108

A particle is moving eastwards with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). In \(5 \mathrm{~s}\) the velocity changes to \(3 \mathrm{~m} / \mathrm{s}\) northwards. The average acceleration in this time interval is (A) \(\frac{1}{2} \mathrm{~m} / \mathrm{s}^{2}\) towards north-east (B) \(1 \mathrm{~m} / \mathrm{s}^{2}\) towards north-west (C) \(\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}\) towards north-east (D) \(\frac{1}{2} \mathrm{~m} / \mathrm{s}^{2}\) towards north-west

Short Answer

Expert verified
The average acceleration is \(\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}\) towards north-east (Answer C).

Step by step solution

01

Determine initial and final velocities

First, we determine the initial and final velocities of the particle. Initial velocity: \(\textbf{v}_i = 4 \mathrm{~m} / \mathrm{s}\) eastwards. Final velocity: \(\textbf{v}_f = 3 \mathrm{~m} / \mathrm{s}\) northwards.
02

Represent velocities as vectors

Now, let's represent the initial and final velocities as vectors. We'll use eastwards as the positive x-direction and northwards as the positive y-direction. Initial velocity vector: \(\textbf{v}_i = 4 \mathrm{~m} / \mathrm{s}\) \(\hat{\imath}\) Final velocity vector: \(\textbf{v}_f = 3 \mathrm{~m} / \mathrm{s}\) \(\hat{\jmath}\)
03

Calculate change in velocity

Next, calculate the change in velocity (\(\Delta\textbf{v}\)) by finding the difference between the initial and final velocity vectors. \(\Delta\textbf{v} = \textbf{v}_f - \textbf{v}_i = (3\hat{\jmath}) - (4\hat{\imath})= -4\hat{\imath} + 3\hat{\jmath}\)
04

Calculate average acceleration

Now, calculate the average acceleration (\(\textbf{a}_{avg}\)) using the formula \(\textbf{a}_{avg} = \frac{\Delta\textbf{v}}{\Delta t}\), where \(\Delta t = 5\mathrm{~s}\). \(\textbf{a}_{avg} = \frac{-4\hat{\imath} + 3\hat{\jmath}}{5 \mathrm{~s}} = - \frac{4}{5}\hat{\imath} + \frac{3}{5}\hat{\jmath}\)
05

Calculate the magnitude and direction of the average acceleration

Finally, find the magnitude and direction of the average acceleration vector. Magnitude: \(|\textbf{a}_{avg}| = \sqrt{(-\frac{4}{5})^2+(\frac{3}{5})^2} = \frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}\) Direction: Using the angle \(\theta\) measured counterclockwise from the East direction (positive x-axis), we have: \(\tan{\theta} = \frac{3/5}{-4/5}\) \(\theta = \arctan{(-3/4)}\) The angle is negative, indicating that the direction is in the south-west quadrant. So the average acceleration is \(\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}\) towards north-east (Answer C).

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