Chapter 3, Problem 1P

A car is driven 225 km west and then 98 km southwest (45º). What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram.

Solution

The magnitude and direction of displacement of the car from the point of origin.

Solution: $302.34\text{km}$ and ${13.2}^{\text{o}}$ South of West

Explanation:

Given:

Initial magnitude of car’s displacement is 225 km in west direction. Then the car turns ${45}^{\text{o}}$ South of west and moves $98\text{km}$ .

Formula used:

The magnitude of the resultant vector and direction can be obtained using the following formulae:

  $\begin{array}{l}R=\sqrt{P^2+Q^2+2PQ\cos \theta }\\ \alpha ={\tan }^{-1}\left(\frac{Q\sin \theta }{P+Q\cos \theta }\right)\end{array}$

Where, P and Q are vectors and $\theta$ is the angle between them.

Calculations:

As per question:

  $\begin{array}{l}P=225\text{km}\\ Q=\text{98}\text{km}\\ \theta ={\text{45}}^{\text{o}}\end{array}$

Substitute the values and solve to find the magnitude of the resultant displacement.

  $\begin{array}{l}R=\sqrt{{225}^2+{98}^2+2\left(225\right)\left(98\right)\cos {45}^{\text{o}}}\\ R=302.34\text{km}\end{array}$

Now, substitute the values to find the direction:

  $\begin{array}{l}\alpha ={\tan }^{-1}\left(\frac{Q\sin \theta }{P+Q\cos \theta }\right)\\ \alpha ={\tan }^{-1}\left(\frac{98\times \sin {45}^{\text{o}}}{225+98\cos {45}^{\text{o}}}\right)\\ \alpha ={13.2}^{\text{o}}\end{array}$

It will be directed South of West.

It can be represented as follows: