Question

A person walks a distance of 30 m towards west with a speed of $2m{s}^{-1}$ and 40 m towards north with a speed of 1. 5 $m{s}^{-1}$ . Then average speed and average velocity for his journey are

• A. $\frac{6}{5}m{s}^{-1},\frac{42}{25}m{s}^{-1}$
• B. $\frac{42}{25}m{s}^{-1},\frac{6}{5}m{s}^{-1}$
• C. $1.68m{s}^{-1},1.19m{s}^{-1}$
• D. $25m{s}^{-1},20m{s}^{-1}$

Answer 1

The correct option is C $1.68m{s}^{-1},1.19m{s}^{-1}$

According to diagram,

Distance

$AB=30m$

$BC=40m$

Velocity

$AB=2m/s$

$BC=1.5m/s$

Now, the distance of AC

$AC=\sqrt{AB+BC}$

$AC=\sqrt{{\left(30\right)}^2+{\left(40\right)}^2}$

$AC=\sqrt{2500}$

$AC=50m$

Now, the time of AB

$t_{AB}=\frac{30}{2}$

$t_{AB}=15\sec$

The time of BC

$t_{BC}=\frac{40}{1.5}$

$t_{BC}=26.67\sec$

Now, the total time $t=26.67sec$

The average speed$\frac{70}{41.67}m/s$

Now,

The average velocity$\frac{50}{41.67}m/s$

Hence, the average speed and velocity are $\frac{70}{41.67}m/s,\frac{50}{41.67}m/s$