Question

A plane flies 470 km east from city A to city B in 45.0 min and then 966 km south from city B to city C in 1.50 h. For the total trip, what are the (a) magnitude and (b) direction of the plane's displacement, the (c) magnitude and (d) direction of its average velocity, and (e) its average speed?

Answer 1

Answer:

a) 1074.27 km

b) $64.06^{\circ}$

c) 477.453 km/h

d) 638.22 km/h

Explanation:

From Pythagoras theorem

AC² = AB²+BC²

$AC=\sqrt{AB^2+BC^2}\\\Rightarrow AC=\sqrt{(-470)^2+(-966)^2}\\\Rightarrow AC=1074.27\ km$

Displacement = 1074.27 km

$tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}\\\Rightarrow tan\theta=\frac{966}{470}\\\Rightarrow \theta=tan^{-1}\frac{-966}{-470}\\\Rightarrow \theta=64.06^{\circ}$

Direction = $64.06^{\circ}$

Average velocity

$v_{av}=\frac{1074.27}{2.25}\\\Rightarrow v_{av}=477.453\ km/h$

Average velocity = 477.453 km/h

Average speed

$s_{av}=\frac{470+966}{2.25}\\\Rightarrow s_{av}=638.22\ km/h$

Average speed = 638.22 km/h