4.5 Relative motion in one and two dimensions (Page 3/8)

University physics volume 1 Page 3 / 8

Flying a plane in a wind

A pilot must fly his plane due north to reach his destination. The plane can fly at 300 km/h in still air. A wind is blowing out of the northeast at 90 km/h. (a) What is the speed of the plane relative to the ground? (b) In what direction must the pilot head her plane to fly due north?

Strategy

The pilot must point her plane somewhat east of north to compensate for the wind velocity. We need to construct a vector equation that contains the velocity of the plane with respect to the ground, the velocity of the plane with respect to the air, and the velocity of the air with respect to the ground. Since these last two quantities are known, we can solve for the velocity of the plane with respect to the ground. We can graph the vectors and use this diagram to evaluate the magnitude of the plane’s velocity with respect to the ground. The diagram will also tell us the angle the plane’s velocity makes with north with respect to the air, which is the direction the pilot must head her plane.

Solution

The vector equation is

{\vec{v}}_{PG} = {\vec{v}}_{PA} + {\vec{v}}_{AG},

where P = plane, A = air, and G = ground. From the geometry in [link] , we can solve easily for the magnitude of the velocity of the plane with respect to the ground and the angle of the plane’s heading,

θ .

A compass shows north is up, east to the right, south down, and west to the left. Vectors V sub P G, V sub A G and V sub P A form a triangle. An airplane is shown on vector V sub P G, which points up. V sub P A points up and to the right, at an angle of theta to vector V sub P G. V sub A G points down and left, at an angle of 45 degrees below the horizontal. V sub P G is the vector sum of v sub P A and V sub A G. — Vector diagram for [link] showing the vectors ${\vec{v}}_{PA}, {\vec{v}}_{AG}, and {\vec{v}}_{PG} .$

(a) Known quantities:

| {\vec{v}}_{PA} | = 300 km / h

| {\vec{v}}_{AG} | = 90 km / h

Substituting into the equation of motion, we obtain $| {\vec{v}}_{PG} | = 230 km / h .$

(b) The angle $θ = {tan}^{−1} \frac{63.64}{300} = 12 °$ east of north.

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Summary

When analyzing motion of an object, the reference frame in terms of position, velocity, and acceleration needs to be specified.
Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies with the choice of reference frame.
If S and $S^{'}$ are two reference frames moving relative to each other at a constant velocity, then the velocity of an object relative to S is equal to its velocity relative to $S^{'}$ plus the velocity of $S^{'}$ relative to S.
If two reference frames are moving relative to each other at a constant velocity, then the accelerations of an object as observed in both reference frames are equal.

Key equations

Position vector	$\vec{r} (t) = x (t) \hat{i} + y (t) \hat{j} + z (t) \hat{k}$
Displacement vector	$Δ \vec{r} = \vec{r} (t_{2}) - \vec{r} (t_{1})$
Velocity vector	$\vec{v} (t) = lim_{Δ t \to 0} \frac{\vec{r} (t + Δ t) - \vec{r} (t)}{Δ t} = \frac{d \vec{r}}{d t}$
Velocity in terms of components	$\vec{v} (t) = v_{x} (t) \hat{i} + v_{y} (t) \hat{j} + v_{z} (t) \hat{k}$
Velocity components	$v_{x} (t) = \frac{d x (t)}{d t} v_{y} (t) = \frac{d y (t)}{d t} v_{z} (t) = \frac{d z (t)}{d t}$
Average velocity	${\vec{v}}_{avg} = \frac{\vec{r} (t_{2}) - \vec{r} (t_{1})}{t_{2} - t_{1}}$
Instantaneous acceleration	$\vec{a} (t) = lim_{t \to 0} \frac{\vec{v} (t + Δ t) - \vec{v} (t)}{Δ t} = \frac{d \vec{v} (t)}{d t}$
Instantaneous acceleration, component form	$\vec{a} (t) = \frac{d v_{x} (t)}{d t} \hat{i} + \frac{d v_{y} (t)}{d t} \hat{j} + \frac{d v_{z} (t)}{d t} \hat{k}$
Instantaneous acceleration as second derivatives of position	$\vec{a} (t) = \frac{d^{2} x (t)}{d t^{2}} \hat{i} + \frac{d^{2} y (t)}{d t^{2}} \hat{j} + \frac{d^{2} z (t)}{d t^{2}} \hat{k}$
Time of flight	$T_{tof} = \frac{2 (v_{0} sin θ)}{g}$
Trajectory	$y = (tan θ_{0}) x - [\frac{g}{2 {(v_{0} cos θ_{0})}^{2}}] x^{2}$
Range	$R = \frac{v_{0}^{2} sin 2 θ_{0}}{g}$
Centripetal acceleration	$a_{C} = \frac{v^{2}}{r}$
Position vector, uniform circular motion	$\vec{r} (t) = A cos ω t \hat{i} + A sin ω t \hat{j}$
Velocity vector, uniform circular motion	$\vec{v} (t) = \frac{d \vec{r} (t)}{d t} = − A ω sin ω t \hat{i} + A ω cos ω t \hat{j}$
Acceleration vector, uniform circular motion	$\vec{a} (t) = \frac{d \vec{v} (t)}{d t} = − A ω^{2} cos ω t \hat{i} - A ω^{2} sin ω t \hat{j}$
Tangential acceleration	$a_{T} = \frac{d \| \vec{v} \|}{d t}$
Total acceleration	$\vec{a} = {\vec{a}}_{C} + {\vec{a}}_{T}$
Position vector in frame S is the position vector in frame $S^{'}$ plus the vector from the origin of S to the origin of $S^{'}$	${\vec{r}}_{P S} = {\vec{r}}_{P S^{'}} + {\vec{r}}_{S^{'} S}$
Relative velocity equation connecting two reference frames	${\vec{v}}_{P S} = {\vec{v}}_{P S^{'}} + {\vec{v}}_{S^{'} S}$
Relative velocity equation connecting more than two reference frames	${\vec{v}}_{P C} = {\vec{v}}_{P A} + {\vec{v}}_{A B} + {\vec{v}}_{B C}$
Relative acceleration equation	${\vec{a}}_{P S} = {\vec{a}}_{P S^{'}} + {\vec{a}}_{S^{'} S}$

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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