PHYS101

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CHAPTER 3: Kinematics in Two or Three Dimensions; Vectors

Suggested problems from Physics for Scientist & Engineers by Giancoli, 4th edition:

Problems: 6, 9, 10, 13, 25, 29, 41, 46, 53, 77, 84, 93 (pages 91-98).

Examples from previous exam questions:

1) (10 points) Two vectors �� and �� are given as, �� = 2𝑖 − 𝑗 and �� = −5𝑖 + 2��. Calculate (�� − 2��).

Answer: �� − 2�� = −9𝑖 + 2𝑗 + 2��

2) (10 points) A ball is shot from the roof of a building into the air with a velocity �� = (2𝑖 + 3𝑗) m/s. In unit vector

notation, what is its velocity 0.5 s later it is shot? (�� = −9.8𝑗 m/s2)

Answer: �� = (2𝑖 − 1.9𝑗) m/s

3) (10 points) Two vectors �� and �� are shown below. Find

a) the angle α,

b) �� + �� in unit vector notation,

Answers: a) α = 60.55°, b) �� + �� = −1.03𝑖 + 3.12𝑗

4) (10 points) For the vectors �� = 3𝑖 + 4𝑗 and �� = 5𝑖 − 2𝑗, give |�� − ��|.

Answer: |�� − ��| = 6.32

5) (10 points) A car initially (𝑡 = 0) located at the origin has an acceleration of (−5.0 𝑚/𝑠2)𝑗 and an initial velocity

of (2.0 𝑚/𝑠)𝑖 . Find the velocity of the car in unit vector notation at 𝑡 = 2 𝑠.

Answer: �� = (2𝑖 − 10𝑗)m/s

6) (10 points) Determine the magnitude of ��2 − 2��1 for vectors ��1 = 12𝑖 − 2𝑗 and ��2 = −2𝑖 + 4𝑗.

Answer: |��2 − 2��1| = 27.2

7) (10 points) A projectile is fired with velocity �� = (12𝑖 + 32𝑗) m/s. Find the time for the projectile to reach the

maximum height. (�� = −9.8𝑗 m/s2)

Answer: t = 3.27 s

52°

|��|=8.4

�� -3.5

-6.2

α

x

y

CHAPTER 3: Kinematics in Two or Three Dimensions; Vectors PHYS101

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8) (10 points) A web page designer creates an animation, in which a dot on the computer screen has a time-dependent

position of 𝑟 = (𝑡2 + 3)𝑖 + (5𝑡)𝑗 where t is in seconds and r is in meters. Find the average velocity of the dot

between t = 0 and t = 2 s.

Answer: ��𝑎𝑣𝑔 = (2𝑖 + 5𝑗) m/s

9) (10 points) If two forces are given in the unit vector notation as ��1 = (5𝑖 − 6𝑗 + 2𝑘) N and ��2 = (3𝑖 + 6𝑗 − 2𝑘)

N, find |��1 − ��2|.

Answer: |��1 − ��2| = 12.8 N

10) (15 points) You are to throw a ball with a speed of v0 =13.0 m/s at a point A on a

wall whose height is h = 5.2 m above the level at which you release the ball. You want

the ball's velocity to be horizontal at the instant it reaches point A.

a) At what angle θ above the horizontal must you throw the ball?

b) What is the horizontal distance from the release point to point A?

Answers: a) θ = 50.95°, b) x = 8.43 m

11) (15 points) Two vectors 𝐴 and �� are shown below. Find

a) the angle α,

b) 𝐴 + �� in unit vector notation,

Answers: a) α = 22°,

b) 𝐴 + �� = 4.50�� − 3.87𝑗 + 3.75��

12) (15 points) At time 𝑡 = 0, a package is released from an airplane

flying horizontally at a height of 500.0 𝑚 above the ground with a

constant speed of 100.0 𝑚/𝑠. Five seconds later, another package is

released.

a) When does the first package hit the ground?

b) What is the velocity vector of the first package just before it

hits the ground?

c) What is the position vector of the second package at the

instant the first one hits the ground?

d) How far away are the packages when both hit the ground?

Answers: a) t = 10.1 s, b) �� = (100𝑖 − 98.99𝑗) m/s, c) 𝑟2 =(1010𝑖 + 372.6��) m, d) ∆𝑥 = 500m

ϑ=100.0 m/s

ground

h=500.0 m

|��|= 10

-9.27

α

y

z

x

40°

𝐴

5.40 -y

-x

CHAPTER 3: Kinematics in Two or Three Dimensions; Vectors PHYS101

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13) (15 points) In Olympic games, an archer fires an arrow horizontally with 96 m/s initial speed to hit the center of

the target (bull’s eye) but the arrow hits a point 47 cm downward.

a) What is the horizontal displacement of the arrow?

b) What should be the initial inclination (angle relative to the horizontal) of the arrow to hit the target from bull’s

eye? [𝑀𝑎𝑡ℎ ℎ𝑒𝑙𝑝: 𝑆𝑖𝑛2𝜃 = 2𝑆𝑖𝑛𝜃𝐶𝑜𝑠𝜃]

Answers: a) x = 29.73 m, b) θ = 0.91°

14) (15 points) The position of an object as a function of time is given by , where t is

in seconds.

a) Find the average velocity of the particle between t = 1.5 s and t = 3.2 s.

b) Determine the object’s instantaneous velocity and instantaneous acceleration as a function of time.

c) What are the magnitudes of the instantaneous velocity and instantaneous acceleration at t = 2.6 s?

Answers: a) ��𝑎𝑣𝑔 = (15.51�� − 8.80��) m/s, b) �� = (6.6𝑡𝑖 − 8.8𝑗) m/s, �� = 6.6𝑖 m/s2,

c) |��| = 19.28 m/s, |��| = 6.6 m/s2

15) (15 points) The position of an object of mass 3 kg is given as a function of time by mkj)62(i6 32 tttr

,

where t is in seconds.

a) Find the average velocity of the particle between t = 2 s and t = 4 s.

b) Determine the object’s instantaneous velocity and acceleration at t = 2 s.

c) Find the net force acting on the object at t = 3 s.

Answers: a) ��𝑎𝑣𝑔 = (36𝑖 − 2𝑗 − 28��) m/s, b) �� = (24𝑖 − 2𝑗 − 12��) m/s, �� = (12𝑖 − 12��) m/s2

c) �� = (36𝑖 − 54��) N

16) (15 points) Figure shows a projectile that is fired with an initial speed of 25 m/s

at an angle of 43° above the horizontal.

a) Find the total horizontal distance it moves when it returns to its initial

height?

b) What is its speed 2.1 s after it is fired?

47 cm

x

25 m/s

43°

y

CHAPTER 3: Kinematics in Two or Three Dimensions; Vectors PHYS101

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c) What is its displacement vector 2.1 s after it is fired?

Answers: a) x = 63.62 m, b) v = 18.62 m/s,

c) ∆𝑟 = (38.4�� + 14.2��) m

17) (15 points) The position of an object is given as a function of time by mk2j)24(i3 32 tttr

, where t is in

seconds.

a) Find the average velocity of the particle between t = 2 s and t = 4 s.

b) Determine the object’s instantaneous velocity and acceleration at t = 2 s.

Answers: a) ��𝑎𝑣𝑔 = (18𝑖 + 4𝑗 − 56��) m/s, b) �� = (12𝑖 + 4𝑗 − 24��) m/s, c) �� = (6𝑖 − 24��) m/s2

18) (15 points) Figure shows a ball that is fired with an initial speed of 40 m/s at an angle of 37° above the horizontal.

The ball hits point A which is horizontally 100 m away as indicated.

a) Find the height of point A.

b) What is its speed when it hits the point A?

c) What is its displacement vector 1.5 s after it is fired?

Answers: a) ℎ = 27.34 m, b) |��| = 32.62 m/s, c) ∆𝑟 = 47.9𝑖 + 25.1𝑗 m