Chapter 3, Problem 41P

(a)

To determine

To find:The velocity of the plane relative to the ground.

Answer to Problem 41P

674.41 km/hr.

Explanation of Solution

Given:

Air plane moving towards south with speed of 600 km/hr and velocity of wind is 100 km/hr.

  $\begin{array}{l}{v}_p=600\text{ km/hr}\\ v_w=100\text{ km/hr}\end{array}$

Formula Used:

  $\begin{array}{l}v=\sqrt{v_x^2+v_y^2}\\ \tan \theta =\frac{v_y}{v_x}\end{array}$

Calculation:

Substituting the given value

  

  $\begin{array}{l}\overrightarrow{v}=50\sqrt{2}\left(-i\right)+600+50\sqrt{2}\left(-j\right)\\ \overrightarrow{v}=70.7\left(-i\right)+670.7\left(-j\right)\\ \overrightarrow{v}=674.41\text{ km/hr}\end{array}$

Conclusion:

Thus, the velocity of the plane to the ground is 674.41 km/hr.

(b)

To determine

The position after 10 mins.

Answer to Problem 41P

16.66 km

Explanation of Solution

Given:

T = 10 mins.

Formula Used:

Relative distance formula

  $R=\frac{d}{t}\times t$

Calculation:

Substitution the value

  $\begin{array}{l}R=\frac{100\text{ km}}{60\text{ min}}\times 10\text{ min}\\ R=16.66\text{ km}\end{array}$

Conclusion:

The position will be 16.66 km after 10 mins.