Question

A jet airliner moving initially at 612 mph (with respect to the ground) to the east moves into a region where the wind is blowing at 362 mph in a direction 15◦ north of east. What is the new speed of the aircraft with respect to the ground? Answer in units of mph.

Answer 1

Answer:

966.22 mph

Explanation:

Velocity of plane with respect to wind (Vp,w)= 612 mph east

velocity of wind with respect to ground, (Vw,g) = 362 mph at 15° North of

east

Write the velocities in vector form

$V_{p,w}=612\widehat{i}$

$V_{w,g}=362\left ( Cos15\widehat{i}+Sin15\widehat{j} \right )= 349.67\widehat{i}+93.69\widehat{j}$

Use the formula for the relative velocity

$V_{p,w}=V_{p,g}-V_{w,g}$

Where, V(p,w) is the velocity of plane with respect to wind

V(p,g) is the velocity of plane with respect to ground

V(w,g) is the velocity of wind with respect to ground

So, $V_{p,g}=V_{p,w}+V_{w,g}$

$V_{p,g}=\left ( 612+349.67 \right )\widehat{i}+93.69\widehat{j}$

$V_{p,g}=961.67\widehat{i}+93.69\widehat{j}$

Magnitude of velocity of lane with respect to ground

$V_{p,g} = \sqrt{961.67^{2}+93.69^{2}}$

V(p,g) = 966.22 mph