Chap 2 & 3: Motion in two or three dimensions

1. A jet plane is flying at a constant altitude. At time t1 = 0 it has components of velocity vx

= 90 m/s, vy = 110 m/s. At time t2 = 30.0 s the components are vx = -170 m/s, vy = 40 m/s
(a) Sketch the velocity vectors at t1 and t2. How do these two vectors differ? For this time
interval calculate
(b) the components of the average acceleration, and
(c) the magnitude and direction of the average acceleration.

IDENTIFY and SET UP: Use Eq. (3.8) in component form to calculate a (av-x) and a(av-y)

EXECUTE: (a) The velocity vectors at t1 = 0 and t2 = 30.0 s are shown in the figure below.

Δv x −170−90
(b) a ( av−x )= = = -8.67 m/s2
Δt 30.0
Δv y 40−110
a ( av− y )= = = -2.33 m/s2
Δt 30.0

(c) a=√ (−8.67)2 +(−2.33)2 = 8.98 m/s2

−2.33
tanθ= = 0.269
−8.67
θ = 15º + 180º = 195º since its in the third quadrant!

2. A web page designer creates an animation in which a dot on a computer screen has a
position of r = [4.0 cm + (2.5 cm/s2)t2] i + (5.0 cm/s)t j.
(a) Find the magnitude and direction of the dot’s average velocity between t = 0 and t =
2.0 s.
(b) Find the magnitude and direction of the instantaneous velocity at t = 0 and t = 1.0 s
and t = 2.0 s.
(c) Sketch the dot’s trajectory from t = 0 to t = 2.0 s, and show the velocities calculated
in part (b).
(a) IDENTIFY and SET UP: From r G we can calculate x and y for any t. Then use
appropriate equations in component form.
EXECUTE: r = [4.0 cm + (2.5 cm/s2)t2] i + (5.0 cm/s)t j
At t = 0, r = (4.0 cm) i
At t = 2.0 s, r = (14.0 cm) i + (10.0 cm) j
Δ x 14.0
v ( av −x )= = = 7.0 cm/s
Δt 2.0
Δ y 10.0
v ( av − y )= = = 5.0 cm/s
Δt 2.0
v (av )=√ (7.0)2 +(5.0)2 = 8.6 cm/s
10.0
tanθ= = 35.5°
14.0

(b) IDENTIFY and SET UP: Calculate r by taking the time derivative of r(t)

EXECUTE:
dr
v= = ([5.0 cm/s2]t)i + (5.0 cm/s)j
dt
At t = 0;
vx = 0, vy = 5.0 cm/s
v=√ (0)2 +(5.0)2 = 5.0 cm/s
5.0
tanθ= =∞
0
θ=90 °

At t = 1.0 s,
vx = 5.0 cm/s, vy = 5.0 cm/s
v=√ (5.0)2 +(5.0)2 = 7.1 cm/s
5.0
tanθ= =1
5.0
θ=45 °

At t = 2.0 s,
vx = 10.0 cm/s, vy = 5.0 cm/s
v=√ (10 .0)2+(5.0)2 = 11 cm/s
5 .0
tanθ= = 0.5
10 .0
 θ=27 °

(c) The trajectory is a graph of y versus x.

x = 4.0 cm + (2.5 cm/s2)t2
y = (5.0 cm/s)t
For values of t between 0 and 2.0 s, calculate x and y and plot y versus x.

3. The position of a squirrel running in a park is given by r = [(0.280 m/s)t + (0.0360 m/s2)t2]
i + (00190 m/s3)t3 j.
(a) What are vx(t) and vy(t), the x- and y-components of the velocity of the squirrel, as
functions of time?
(b) At t = 5.00 s, how far is the squirrel from its initial position?
(c) At t = 5.00 s, what are the magnitude and direction of the squirrel’s velocity?

IDENTIFY:
Given the position vector of a squirrel, find its velocity components in general, and at a
specific time find its velocity components and the magnitude and direction of its
position vector and velocity.

SET UP: vx = dx/dt and vy = dy/dt; the magnitude of a vector is A=√ Ax 2+ Ay 2

 EXECUTE: (a) Taking the derivatives gives vx(t) = 0.280 m/s + (0 0720 m/s2)t and vy(t) =
(0.0570 m/s3).
(b) Evaluating the position vector at t = 5.00 s gives x = 2.30 m and y = 2.375 m, which
gives r = 3.31 m.
(c) At t = 5.00 s, vx = +0.64 m/s, vy = 1.425 m/s, which gives v = 1.56 m/s and
tanθ=1.425 /0.64 so the direction is θ = 65.8° (counterclockwise from +x-axis)

4. A dog running in an open field has components of velocity vx = 2.6 m/s and vy = -1.8 m/s
at t1 = 10.0 s. For the time interval from t1 = 10.0 s to t2 = 20.0 s, the average acceleration
of the dog has magnitude 0.45 m/s2 and direction 31.0° measured from the +x-axis
toward the +y-axis. At t2 = 20.0 s,
(a) what are the x- and y-components of the dog’s velocity?
(b) What are the magnitude and direction of the dog’s velocity?
(c) Sketch the velocity vectors at t1 and t2 and how do these two vectors differ?

SET UP:
ax = (0.45 m/s2)cos31.0 ° = 0.39 m/s2
ay = (0.45 m/s2)sin31.0 ° = 0.23 m/s2
Δv x
(a) a ( av−x )= and vx = 2.6 m/s + (0.39 m/s2)(10.0 s) = 6.5 m/s
Δt
a ¿ and vy = -1.8 m/s + (0.23 m/s2)(10.0 s) = 0.52 m/s
(b) v=√(6.5)2 +(0.52)2 = 6.52 m/s
0.52
tanθ=
6.5
θ=4.6 ° above the horizontal
(c) ) The velocity vectors v1 and v2 are sketched below. The two velocity vectors differ in
magnitude and direction.

v1 is at an angle of 35° below the +x-axis and has magnitude v1 = 3.2 m/s, so v1 > v2 and the
direction of v2 is rotated counterclockwise from the direction of v1
 5. The coordinates of a bird flying in the xy-plane are given by x(t) = αt and y(t) = 3.0 m –
βt2 where α = 2.4 and β = 1.2 m/s
(a) Sketch the path of the bird between t = 0 and t = 2.0 s
(b) Calculate the velocity and acceleration vectors of the bird as functions of time.
(c) Calculate the magnitude and direction of the bird’s velocity and acceleration at t =
2.0 s
(d) Sketch the velocity and acceleration vectors at t = 2.0 s. At this instant, is the bird
speeding up, is it slowing down, or is its speed instantaneously not changing? Is the bird
turning? If so, in what direction?

IDENTIFY and SET UP: Use the appropriate equations to find vx , vy, ax and ay as functions of
time. The magnitude and direction of r and a can be found once we know their components.

EXECUTE:
(a) Calculate x and y for t values in the range 0 to 2.0 s and plot y versus x. The results are
given in the diagram below.

dx
(b) v ( x )= =α
dt
dy
v ( y )= =−2 βt
dt
d v (x)
a ( x )= =0
dt
d v ( y)
a ( y )= =−2 β
dt
Thus, v=α i−2 βt j
a=−2 β j
 (c) velocity; At t = 2.00 s,
v(x) = 2.4 m/s
v ( y)=−2(1.2)(2.0) = -4.8 m/s

v=√ (2.4)2 +(−4.8)2 = 5.4 m/s

−4.8
tan α = = -2.00
2.4
α =−63.4 °+360 ° = 297°, since its in the 4th quadrant.

Acceleration; At t = 2.00 s,
a(x) = 0
a ( y)=−2(1.2) = -2.4 m/s2
a=√ (0)2 +(−2.4)2 = 2.4 m/s2
−2.4
tan β = = -∞
0
β = 270°

(d)

a has a component a|| in the same direction as v so we know that v is increasing (the bird is
speeding up.) a also has a component a⊥ perpendicular to v so that the direction of v is
changing; the bird is turning toward the -direction −y (toward the right).
v is always tangent to the path; v at t = 2.00 s shown in part (c) is tangent to the path at this t,
conforming to this general rule. a is constant and in the −y-direction; the direction of v is
turning toward the −y-direction.

Chap 1: Vectors

1. You are hungry and decide to go to your favorite neighborhood fast-food restaurant.
You leave your apartment and take the elevator 10 flights down (each flight is 3.0 m)
and then go 15 m south to the apartment exit. You then proceed 0.2 km east, turn
north, and go 0.1 km to the entrance of the restaurant.
 (a) Determine the displacement from your apartment to the restaurant. Use unit vector
notation for your answer, being sure to make clear your choice of coordinates.
(b) How far did you travel along the path you took from your apartment to the
restaurant, and what is the magnitude of the displacement you calculated in part (a)?
ANSWER:
IDENTIFY. Find the vector sum of the four displacements.

SET-UP: Take the beginning of the journey as the origin, with north being the y-direction, east
the x-direction, and the z-axis vertical. The first displacement is then (-30 m) k, the second is (-
15 m) j, the third is (200 m) i, and the fourth is (100 m) j.

EXECUTE: (a) Adding the four displacements gives (-30 m) k + (-15 m) j + (200 m) i + (100 m) j =
(200 m) i + (85 m) j - (30 m) k.
(b) The total distance traveled is the sum of the distances of the individual segments: 30 m + 15
m + 200 m + 100 m = 345 m. The magnitude of the total displacement is:
D= √ Dx 2+ Dy 2+ Dz2=√ (200 m)2 +(85 m)2+(−30 m)2 = 219 m.

2. A fence post is 52.0 m from where you are standing, in a direction north of east. A
second fence post is due south from you. What is the distance of the second post from
you, if the distance between the two posts is 80.0 m?
ANSWER:
IDENTIFY: The displacements are vectors in which we know the magnitude of the resultant and
want to find the magnitude of one of the other vectors.

SET UP: Calling A the vector from you to the first post, B the vector from you to the second
post, and C the vector from the first to the second post, we have A +B + C. Solving using
components and the magnitude of C gives Ax + Cx = Bx and Ay + Cy = By

EXECUTE: Bx = 0, Ax = 41.53 m and Cx = Bx - Ax = - 41.53 m.

C = 80.0 m, so
Cy=± √ C 2−Cx 2=± 68.38 m.

The post is 37.1 m from you.

EVALUATE: By = −37.1 m (negative) since post is south of you (in the negative y direction).

3. Ricardo and Jane are standing under a tree in the middle of a pasture. An argument
ensues, and they walk away in different directions. Ricardo walks 26.0 m in a direction
 west of north. Jane walks 16.0 m in a direction south of west. They then stop and turn to
face each other.
(a) What is the distance between them?
(b) In what direction should Ricardo walk to go directly toward Jane?

IDENTIFY: The displacements are vectors in which we know the magnitude of the resultant and
want to find the magnitude of one of the other vectors.

SET UP: Calling A the vector of Ricardo’s displacement from the tree, B the vector of Jane’s
displacement from the tree, and C the vector from Ricardo to Jane, we have A + C = B. Solving
using components we have Ax + Cx = Bx and Ay + Cy = By

EXECUTE:
(a) The components of A and B are
Ax = -(26.0 m) sin60.0 = -22.52 m
Ay = (26.0 m) cos60.0 = 13.0 m
Bx = -(16.0 m) cos30.0 = -13.86 m
By = -(16.0 m) sin30.0 = -8.00 m
Cx = Bx – Ax = -13.86 m - (-22.52 m) = 8.66 m
Cy = By – Ay = -8.00 m - (13.0 m) = -21.0 m
Finding the magnitude from the components gives C = 22.7 m.
(b) Finding the direction from the components gives
8.66
tanθ= = 22.4 °, east of south.
21.0

4. Given two vectors A = 4.00i + 7.00j and B = 5.00i – 2.00j.

(a) find the magnitude of each vector;
(b) write an expression for the vector difference A - B using unit vectors;
(c) find the magnitude and direction of the vector difference A - B
(d) In a vector diagram show A, B and A -B, and also show that your diagram agrees
qualitatively with your answer in part (c).

IDENTIFY: Find A and B. Find the vector difference using components.

SET UP: Deduce the x- and y-components and use appropriate equations

EVALUTE:
EXECUTE:
 (a) A = 4.00 i + 7.00 j
Ax = +4.00
Ay = +7.00

A=√ Ax 2+ Ay 2=√ (4.00)2+(7.00)2=8.06

B = 5.00 i - 2.00 j
Bx = +5.00
By = -2.00

B=√ B x 2 + B y 2=√ (5 .00)2+(−2 .00)2=5.39

(b) A – B = 4.00 i + 7.00 j - (5.00 i – 2.00 j) = (4.00 – 5.00) i + (7.00 – 2.00) j

A – B = -1.00 i + 9.00 j

(c) Let A – B = R
Then R = -1.00 i + 9.00 j
Rx = -1.00
Ry = 9.00

R=√ (−1.00)2+(9.00)2 = 9.06

9.00
tanθ= = -9.00
−1.00
θ=−83.6 ° +180 °=96.3 ° , since its in the second quadrant.

5. A plane leaves the airport in Galisteo and flies 170 km at of north and then changes
direction to fly 230 km at of east, after which it makes an immediate emergency landing
in a pasture. When the airport sends out a rescue crew, in which direction and how far
should this crew fly to go directly to this plane?

IDENTIFY: Find the vector sum of the two displacements.

SET UP: Call the two displacements A and B where A = 170 km and b = 230 km. A + B = R.
A and B are as shown in the diagram below

EXECUTE:
Rx = Ax + Bx = (170 km) sin68 + (230 km) cos48 = 311.5 km.
Ry = Ay + By = (170 km) cos68 - (230 km) sin48 = -107.2 km.
R=√ (311.5)2 +(−107.2)¿2 ¿ = 330 km.
 107.2
tanθ= = 19° south of east.
−311.5