(a) Another airplane is flying in a jet stream that is blowing at m/sin a direction south of east. Its direction of motion relative to the Earth is south of west, while its direction of travel relative to the air is south of west. What is the airplane’s speed relative to the air mass?

(b) What is the airplane’s speed relative to the Earth?

Short Answer

Expert verified

The velocity of the plane with respect to air is 63.5 m/s. The velocity of the plane relative to the earth is 29.6 m/s.

Step by step solution

01

Given data

The velocity of air relative to the earth is 45 m/s. It is in 20 degree.

The sum of the plane's velocity relative to the air and the air's velocity relative to the earth equals the plane's velocity relative to the earth.

02

Velocity of the plane relative to the earth

The velocity of the plane with respect to the earth:

Vpe=Vpa+Vae

The component table will be as below:

X

Y

Vpa

-Vpacos5°

-Vpasin5°

Vae

42.3

-15.4

Resultant

-Vpecos45°

-Vpesin45°

Here we need to find the x component and the y component.

XcomponentVx=VicosθVx=-VpacosθVx=-Vpacos5°

YcomponentVy=VisinθVy=-VpasinθVy=-Vpasin5°

Hence, the velocity of air w.r.t to the earth component :

XComponentVx=VicosθVx=45cos20°Vx=42.3

YComponentVy=VisinθVy=-45sin40°Vy=-15.4

03

Angle of the resultant vectors

Now we have given the angle of the resultant vector; hence

XcomponentVx=VpecosθVpex=-Vpecos45°YcomponentVpey=VisinθVy=-Vpesin45°

Hence, putting the value in the equation

-Vpacos45°+42.3=-Vpecos45°......(1)Andequationtois:-Vpasin45°+-15.4=-Vpesin45°......(2)

After solving the above two equations, we can get the value of Vpa and Vpe.

Hence, Vpa= 63.5 m/s.

The velocity of the plane with respect to air is 63.5 m/s.

04

Speed of plane relative to the earth

b) The speed of the plane relative to the earth should be

Vpe=Vpasin5°+15.4sin45°Vpe=29.6ms

The velocity of the plane relative to the earth is 29.6 m/s.

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Most popular questions from this chapter

In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77m. What was the initial speed of the shot if he released it at a height of2.10m and 38.0°threw it at an angle of above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45° when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus,38° will give a longer range than45° in the shot put.)

Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. How far are you from your starting point and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure 3.56, then this problem finds their sum R = A + B.)

Find the following for path A in Figure,

(a) the total distance traveled, and

(b) the magnitude and direction of the displacement from start to finish.

The various lines represent paths taken by different people walking in a city. All blocks are120 mon a side.

(a) Repeat the problem two problems prior, but for the second leg you walk \(20.0{\rm{ m}}\) in a direction \(40.0^\circ \) north of east (which is equivalent to subtracting \({\rm{B}}\) from \({\rm{A}}\) —that is, to finding \({\rm{R'}} = {\rm{A}} - {\rm{B}}\)).

(b) Repeat the problem two problems prior, but now you first walk \(20.0{\rm{ m}}\) in a direction \(40.0^\circ \) south of west and then \(12.0{\rm{ m}}\) in a direction \(20.0^\circ \) east of south (which is equivalent to subtracting \({\rm{A}}\) from \({\rm{B}}\) —that is, to finding \({\rm{R''}} = {\rm{B}} - {\rm{A}} = - {\rm{R'}}\)). Show that this is the case.

A ball is kicked with an initial velocity of \(16\,m/s\) in the horizontal direction and \(12\,m/s\) in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c) What maximum height is attained by the ball?

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