Question

An airplane maintains a speed of630 km/hrelative to the air it is flying through as it makes a trip to a city750 kmaway to the north.

(a) What time interval is required for the trip if the plane flies through a headwind blowing at 35.0 km/htoward the south?

(b) What time interval is required if there is a tailwind with the same speed?

(c) What time interval is required if there is a crosswind blowing at35.0 km/hto the east relative to the ground?

Short answer

(a) The time interval required for the trip is,1.26 h.

(b) The time interval required for the trip if the plane flies through the tail wind is1.13 h.

(c) The time interval required for the trip if there is a crosswind is 1.19 h.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The speed of the airplane is $630 \text{km/h}$.
• The distance travelled by the airplane is 750 km.
• The speed of the wind is$35.0 \text{km/h}$ .

The time interval required for the trip

The time interval required for the trip is,

$\mathit{t}\mathbf{=}\frac{\mathbf{d}}{\left(v_p-v_w\right)}$

Here,

v_p is the velocity of the plane.

v_w is the velocity of the wind.

d is the distance travelled by the airplane.

Determination of the time interval is required for the trip if the plane flies

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}t& =& \frac{750 \text{km}}{(630 \text{km/h}-35.0 \text{km/h})}\\ & =& 1.26 \text{h}\\ & & \end{array}$

Therefore, the time interval required for the trip is 1.26 h.

The time interval required for the trip if the plane flies through the tail wind

(b)

The formula to calculate the time interval required for the trip if the airplane flies through tail wind is,

$t=\frac{d}{\left(v_{\text{p}}+v_{\text{w}}\right)}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}t& =& \frac{750 \text{km}}{(630 \text{km/h}+35.0 \text{km/h})}\\ & =& 1.127 \text{h}\\ & \cong & 1.13 \text{h}\\ & & \end{array}$

Therefore, the time interval required for the trip if the plane flies through the tail wind is $\begin{array}{rcl}& & 1.13 \text{h}\end{array}$.

The time interval required for the trip if there is a crosswind

(c)

The cross wind is blowing to the east relative to the ground, due to this airplane instead of heading exact north, the airplane will move towards west of north. In order to make the airplane head exact north, the airplane has to counter the angle change due to cross wind.

Write the expression to calculate the velocity of the airplane due east.

$v_{\text{p,east}}^{\text{'}}=v_P\sin \theta +\left(-v_w\right)$

Here, $v_{\text{p,east}}^{\text{'}}$ is the velocity of the airplane along east with respect to ground and is the angle $\theta$To counter this effect of the cross wind role="math" localid="1663800909768" $v_{\text{p,east}}^{\text{'}}=0$,

Substitute the above expression, and we get,

$\begin{array}{rcl}0& =& (630 \text{km/h})\sin \theta -35.0 \text{km/h}\\ \sin \theta & =& \frac{35.0 \text{km/h}}{630 \text{km/h}}\\ \theta & =& {\sin }^{-1}\left(\frac{35.0 \text{km/h}}{630 \text{km/h}}\right)\\ & =& 3.18°\\ & & \end{array}$

The expression for the velocity of airplane due north, $v_{\text{p},\text{north}}^{\text{'}}=v_P\cos \theta$

The cross wind is blowing to the east relative to the ground. Hence, it will not have any effect along velocity of airplane due north.

Substitute the above expression, and we get,

$\begin{array}{rcl}{v}_{\text{p},\text{north}}^{\text{'}}& =& (630 \text{km/h})\cos \left(3.18°\right)\\ & =& 629 \text{km/h}\\ & & \end{array}$

The time interval required for the trip is,

$t=\frac{d}{v_{\text{p}}}$

Substitute the above expression, and we get,

$\begin{array}{rcl}t& =& \frac{750 \text{km}}{629 \text{km/h}}\\ & =& 1.19 \text{h}\\ & & \end{array}$

Therefore, the time interval required for the trip if there is a crosswind is$\begin{array}{rcl}& & 1.19 \text{h}\end{array}$ .