Question

A train moves at a constant speed of 60.0 km/h towards east for 30 min, then in a direction ${30}^{\circ }$ east of due north for 20.0 min, and then west for 45 min. What are the (a) magnitude(in km/min) and (b) angle of its average velocity during this trip?

• A. Magnitude=$\left(\frac{19}{\sqrt{13}}\right);ta{n}^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
• B. Magnitude= $\frac{\sqrt{13}}{19};ta{n}^{-1}(-2\sqrt{3})$
• C. Magnitude=$\left(\frac{\sqrt{13}}{19}\right);ta{n}^{-1}\left(2\sqrt{3}\right)$
• D. Magnitude=$\frac{19}{\sqrt{13}};ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer 1

The correct option is B

Magnitude= $\frac{\sqrt{13}}{19};ta{n}^{-1}(-2\sqrt{3})$

So we need to find the total displacement
Let's draw the displacement vector and see but how much is the body displaced at the end of the motion.

So writing all the three displacement vectors of the three parts of motion - $30\hat{i};10\hat{i}+10\sqrt{3}\hat{j};-45\hat{i}$
Simply add all three to get final position.
Final position is $-5\hat{i}+10\sqrt{3}\hat{j}$
So the particle was displaced from $0\hat{i}+0\hat{j}to-5\hat{i}+10\sqrt{3}\hat{j}$ in time
(30 min + 20 min + 45 min)
=95 min
Magnitude of displacement =$\sqrt{25+300}=\sqrt{325}=5\sqrt{13}$
Direction =$ta{n}^{-1}\left(\frac{10\sqrt{3}}{-5}\right)$
$=ta{n}^{-1}(-2\sqrt{3})$
Average velocity = $\frac{displacement}{time}=\frac{5\sqrt{13}}{95}=\frac{\sqrt{13}}{19}$
Direction of displacement = Direction of average velocity -$ta{n}^{-1}(-2\sqrt{3})$