Kinematics

 Study that deals with the description of motion

 Distance vs displacement
o On his way to SM, Patrick traveled 100 m north, 300 m east, 200 m North, 200 m east, and
finally 300 m north. Find (a) the total distance traveled by Patrick and (b) the displacement
made by Patrick.
 Speed and velocity
dis tan ce(m)
o vs =
time( s )
vi  vf
o vs =
2
displacement (m)
o v=
time( s )
d 2  d1
o vs =
t 2  t1
 acceleration
vf  vi
o a=
tf  ti
 kinematic equations
d
o v=
t
vf  vi
o v=
Examples 2
1. How vf  vi
o far
a = does a jogger run in 1.5 hours if his average speed is 2.22 m/s?
2. A plane starts t from rest and accelerates down the runway and at 29 s attains a velocity of 260
km/hr.  vf  vi the acceleration of the plane.
o d =Determine
 t d = ½ (vf + vi)t
2  the finish line, and the driver deploys a parachute and applies the brakes to
 crosses
3. A drag racer
slow down. The atdriver
2
begins slowing down when ti = 9.0 s and the car’s velocity is vi = 28 m/s.
o d = vit +
When tf = 12.0 s, 2 the velocity has been reduced to vf = 13 m/s. What is the average acceleration
of the dragster?
vf  vi
2 2

4. The d=
o speedboat
2ahas a constant acceleration of 2.0 m/s . If the initial velocity of the boat is 6.0
2

m/s, find its displacement after 8.0 seconds.

5. A jet is taking off from the deck of an aircraft carrier. Starting from rest, the jest is
catapulted with a constant acceleration of 31 m/s 2 along a straight line and reaches a velocity of
62 m/s. Find the displacement of the jet.
6. A motorcycle starting from rest has an acceleration of 2.6 m/s 2. After the motorcycle has
traveled a distance of 120 m, it slows down with an acceleration of -1.5 m/s 2 until its velocity is 12
m/s. What is the total displacement of the motorcycle?
7. A runner accelerates to a velocity of 5.36 m/s due west in 3 s. His average acceleration is 0.640
m/s2, also directed to the west. What was his velocity when he began accelerating?
8. The 1999 VW Beetle goes from 0 to 60.0 mi/h with an acceleration of 2.35 m/s 2. (a) How much
time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go fro 0 to 60.0
mi/h in 0.600 s. What is the acceleration (in m/s) of the dragster?
9. In getting ready to slam-duck the ball, a basketball player starts from rest and sprints to a speed
of 6.0 m/s in 1.5 s. Assuming that the player accelerates uniformly, determine the distance he
runs.
10. A car is traveling on a dry road with a velocity of 32 m/s. The driver slams on the brakes and
skids to a halt with an acceleration of -8.0 m/s. On an icy road, the car would have skidded to a
halt with an acceleration of -3.0 m/s. How much farther would the car have skidded on the icy
road compared to the dry road? 30 Dorothy D. Silva 
Solutions:
1. Given: speed = 2.22 m/s Required: distance
time = 5400 s
Solution:
Distance = (speed) (time)
= (2.22 m/s) (5400 s)
Distance = 11, 988 m

2. Given: vf = (260 km/hr) 72.22 m/s Required: acceleration

vi = 0 m/s
t = 29 s
Solution:
vf  vi
a=
tf  ti
72.22m / s  0m / s
a= = 2.49 m/s2
29 s  0 s

3. Given: vf = 13 m/s Required: acceleration

vi = 28 m/s

31 Dorothy D. Silva 
 tf = 12 s
ti = 9 s
Solution:
vf  vi
a=
tf  ti
13m / s  28m / s
a=
12 s  9 s
a = -5.00 m/s2 (decelerating)

4. Given: a = 2 m/s2 Required: diplacement

vi = 6 m/s vf
t=8s

Solution:
vf = vi + at
vf = 6 m/s + (2 m/s2 x 8 s) = 22.00 m/s
d = ½ (vi + vf) t
d = ½ (6 m/s + 22 m/s) (8 s) = 112.00 m

5. Given: a = 31 m/s2 Required: diplacement

vi = 0 m/s t
vf = 62 m/s
t=8s

Solution:
vf  vi
t=
a
62m / s  0m / s
t= = 2.00 s
31m / s 2
d = ½ (vi + vf) t
d = ½ (0 m/s + 62 m/s) (2 s) = 62.00 m

6. Given: a = 2.6 m/s2 a = -1.5 m/s2

vi = 0 m/s vf = 12 m/s
d = 120 m vi = ?
vf = ?

Solution:
vf = vi 2  2ad
vf = (0m / s ) 2  2( 2.6m / s 2 )(120m) = 24.98 m/s
 vi 2
vf 2
d=
2a
(12m / s ) 2  ( 25m / s ) 2
d= = 160.33 m
(2)(1.5m / s 2 )
dtotal = 120 m + 160.33 m
dtotal = 280.33 m

7. Given: a = 0.640 m/s2 vi = ?

32 Dorothy D. Silva 
 vf = 5.36 m/s
t=3s

Solution:
vi = vf - at
vi = (5.36 m/s) – (0.640 m/s2)(3 s)
vi = 3.44 m/s

8. Given: a = 2.35 m/s2 vi = 0 m/s

vi = 0 m/s vf = 60.0 mi/hr = 26.82 m/s
vf = 60.0 m/s = 26.82 m/s t = 0.600 s
t=? a=?

Solution:
vf  vi
t=
a
26.82m / s  0m / s
t= = 11.41 s
2.35m / s 2

vf  vi
a=
t
26.82m / s  0m / s
a= = 44.70 m/s2
0.600 s

9. Given: vi = 0 m/s d=?

vf = 6.0 m/s
t = 15 s

Solution:
 vf  vi 
d=  t
 2 
 6.0m / s  0m / s 
d=  15s
 2 
d = 45 m

10. Given: a = -8.0 m/s2 a = -3.0 m/s2

vi = 32 m/s vi = 32 m/s
vf = 0 m/s vf = 0 m/s
d=? d=?

33 Dorothy D. Silva 
 Solution:
vf vi 2
2
vf vi 2
2
d= d=
2a 2a
(0m / s) 2  (32m / s) 2 (0m / s) 2  (32m / s) 2
d= d=
(2)(8m / s 2 ) (2)(3m / s 2 )
d = 64 m d = 170.67 m

Graphical Analysis of motion

 distance vs time graph
 velocity vs time graph

(a)+ Velocity Constant Velocity (b) + Velocity Changing Velocity

(a) Slow, Rightward (+) (b) Fast, Rightward (+)

o
(a)Slow, Leftward (–) (b)Fast, Leftward (–)

(a)Leftward (–) Velocity (b)Leftward (–) Velocity

Free Falling Bodies


g = 9.8 m/s2 or 32.2 ft/s2
o Christian Huygens (pendulum clock)

Galileo

David Scott (Apollo 15)
34 Dorothy D. Silva 
 d
o v=
t
o vf = vi + gt
 vf  vi 
o d=  t
 2 
gt 2
o d = vit +
2
vf 2
 vi 2
o d=
2g

Examples:
1. A stone is dropped from rest from the top of a tall building. After 3.0 s of free-fall,
what is the displacement of the stone? After 3.0 s of free-fall, what is the velocity of
the stone?
2. A football game customarily begins with a coin toss to determine who kicks off. The
referee tosses the coin up with an initial speed of 6.0 m/s. In the absence of air
resistance, how high does the coin go above its point of release? What is the total time
the coin is in the air before returning to its release point?
3. The greatest height reported for a jump into airbag is 99.4 m by a stuntman Dan Koko.
In 1948, he jumped from rest from the top of the Vegas World Hotel and Casino. He
struck the airbag at a speed of 39 m/s. To assess the effects of air resistance,
determine how fast he would have been traveling on impact had air resistance been
absent.
4. An astronaut on a distant planet wants to determine its acceleration due to gravity.
The astronaut throws a rock straight up with a velocity of 15 m/s and measures a time
of 20.0 s before the rock returns to his hand. What is the acceleration due to gravity
on this planet?
5. From her bedroom window a girl drops a water-filled balloon to the ground 6.0 m below.
If the balloon is released from rest, how long is it in air?
6. A wrecking ball is hanging at rest from a crane when suddenly the cable breaks. The
time it takes for the ball to fall halfway the ground is 1.2 s. Find the time it takes for
the ball to fall from rest all the way to the ground.
7. Two arrows are shot vertically upward. The second arrow is shot after the first one,
but while the first is still on its way up. The initial speed are such that both arrows
reach their maximum height at the same instant, although these heights are different.
Suppose that the initial speed of the first arrow is 25.0 m/s and that the second arrow
is fired 1.20 s after the first. Determine the initial velocity of the second arrow.

Solutions:
1. Given: g = -9.80 m/s2 d=?
vi = 0 m/s vf = ?
t = 3.00 s

35 Dorothy D. Silva 
 Solution:
gt 2
d = vit +
2
( 9.80m / s 2 )(3.00 s ) 2
d = (0 m/s)(3.00 s) +
2
d = -44.10 m (44.10 m downward)

vf = vi + gt
vf = 0 m/s + (-9.80 m/s2)(3.00 s)
vf = -29.40 m/s (29.40 m/s downward)

2. Given: g = -9.8.0 m/s2 d=?

vi = 0 m/s t=?
vf = 6.00 m/s

Solution:
vf 2
 vi 2
d=
2g
(0m / s ) 2  (6.00m / s ) 2
d=
(2)(9.80m / s 2 )
d = 1.84 m

vf  vi
t=
g
0m / s  6.00m / s
t=
 9.80m / s 2
t = 0.61 s

ttotal = (0.61 s) (2) = 1.22 s

3. Given: g = -9.8.0 m/s2 vf = ?

vi = 0 m/s
d = 99.4 m

Solution:
vf = 2 gd  vi 2
vf = 2( 9.80m / s 2 )(99.4m)  (0m / s ) 2
vf = 44.14 m/s

4. Given: vi = 15 m/s g=?

vf = 0 m/s
ttotal = 20.0 s
t1 = t2 = 10.0 s

Solution:

36 Dorothy D. Silva 
 vf  vi
g=
t
0m / s  15m / s
g=
10 s
g = -1.5 m/s2 (1.5 m/s2 downward)

5. Given: g = -9.8.0 m/s2 vf = ?

vi = 0 m/s t=?
d = -6.0 m

Solution:
vf = 2 gd  vi 2
vf = 2( 9.80m / s 2 )(6.0m)  (0m / s ) 2
vf = 10.84 m/s downward

vf  vi
t=
g
 10.84m / s  0m / s
t=
 9.80m / s 2
t = 1.11 s

6. Given: g = -9.8.0 m/s2 vf = ?

vi = 0 m/s ttotal = ?
t = 1.20 s (halfway)

Solution:
gt 2
d = vit +
2
(9.80m / s 2 )(1.20 s ) 2
d = (0 m/s) (1.20s) +
2
d = 7.06 m
dtotal = 7.06 m x 2 = 14.12 m

vf = 2 gd  vi 2
vf = 2( 9.80m / s 2 )(14.12m)  (0m / s ) 2
vf = 16.64 m/s (downward)

vf  vi
t=
g
 16.64m / s  0m / s
t=
 9.80m / s 2
t = 1.70 s

7. Given: g = -9.8.0 m/s2 d=?

vi = 25 m/s t=?
vf = 0 m/s

37 Dorothy D. Silva 
 Solution:
first arrow
vf 2
 vi 2
d=
2g
(0m / s ) 2  (25m / s ) 2
d=
( 2)(9.80m / s 2 )
d = 31.89 m

vf  vi
t=
g
0m / s  25.00m / s
t=
 9.80m / s 2
t1 = 2.55 s

second arrow
t2 = 2.55 – 1.20
t2 = 1.35 s

vi = vf – gt
vi = 0 m/s – (-9.8 m/s2)(1.35 s)
vi = 13.23 m/s

Projectile Motion
1. A kind of two-dimensional motion that occurs when the moving object (projectile) experiences only
the acceleration due to gravity, which acts in the vertical direction
a. Trajectory – curved path
b. Projectile – an abject thrown with an initial horizontal velocity and acted upon by the earth’s
pull of gravity

A. Horizontal Projection
 Projected horizontally
 Parallel to a level surface

38 Dorothy D. Silva 
  Horizontal velocity remains constant
o Time of flight
o Length of the horizontal component of velocity
 ax = 0
 ay = g
 Vx = constant
 Vy = gt
 dy = gt2/2
 Vy = 2 gdy
 dx = Vxt
 dx = range

Examples:
1. A ball projected from a height of 25.0 m above the ground and is thrown with an initial velocity
of 8.25 m/s.
a. How long is the ball in flight before striking the ground?
b. How far from the building does the ball strike the ground?

2. A little girl throws her jackstone ball horizontally out of the window with a velocity of 30 m/s.
If the window is 3 m above the level ground, how far will the ball travel before it hits the
ground?

3. A diver runs horizontally with a speed of 1.20 m/s off a platform that is 10.0 m above the
water. What is his speed just before striking the water?

4. An airplane moving horizontally with a constant velocity of 115 m/s at an altitude of 1050 m.
The plane releases a “care package” that falls to the ground along a curved trajectory. Ignoring
air resistance, (a) determine the time required for the package to hit the ground. Suppose that
the plane is traveling with twice the horizontal velocity (230 m/s), (b) determine the time
required for the package to hit the ground.

Solutions:
1. Given: g = -9.8.0 m/s2 a) t = ?
vx = 8.25 m/s b) dx = ?
dy = 25.0 m
Solution:
2dy
a. t =
g
2( 25m / s )
t= t = 2.26 s
 9.8m / s 2

39 Dorothy D. Silva 
 b. dx = (vx)(t)
dx = (8.25 m/s) (2.26 s)
dx = 18.65 m

2. Given: g = -9.8.0 m/s2 dx = ?

vx = 30 m/s
dy = 3 m
Solution:
2dy
t=
g
2(3m / s )
t= t = 0.78 s
 9.8m / s 2

dx = (vx)(t)
dx = (30 m/s) (0.78 s)
dx = 23.40 m

3. Given: g = -9.8.0 m/s2 vy = ?

vx = 1.20 m/s
dy = 10.0 m
Solution:
vy = 2 gdy
vy = 2( 9.8m / s 2 )(10 m)
vy = 14 m/s

4. Given: g = -9.8.0 m/s2 t=? g = -9.80 m/s2

vx = 115 m/s vx = 230 m/s
dy = 1050 m dy = 1050 m
Solution:
2dy 2dy
t= t=
g g
2(1050m / s ) 2(1050m / s )
t= t=
 9.8m / s 2  9.8m / s 2
t = 14.64 s t = 14.64 s

B. Projections at Arbitrary Angles


Launch speed = landing speed
o Y initial = Y final

Parabola – curved produced by projection at an angle
o Parabolic

Θ (launch point) = θ (target point)

vix = vi cos θ

viy = vi sin θ
gt 2

dy = viyt +
2

dy = vfy2 – Viy2

40 Dorothy D. Silva 
 2g

vfy = viy + gt
2viy viy

ttotal = t=
g g
2vi sin  vi sin 

ttotal = t=
g g

dy = (vi sin θ)2 R = (vi2 sin 2θ) R = (vix) (t)
2g g

Examples:
1. A place-kicker kicks a football at an angle of 40° above the horizontal axis. The speed of the
ball is vi = 22 m/s. Find the maximum height the ball attains.

2. Suppose a golf ball is hit off the tee with an initial velocity of 30.0 m/s at an angle of 35° to
the horizontal.
a. What is the maximum height reached by the ball?
b. What is the range?

3. A rifle fires a bullet with a speed of 250 m/s at an angle of 37° above the horizontal.
a. What height does the bullet reach?
b. How long is the bullet in the air?
c. What is the horizontal range?

4. An arrow has an initial launch speed of 18 m/s. If it must strike a target 31 m away at the same
elevation, what should be the projection angle?

Solutions:
1. Given: g = -9.8.0 m/s2 dy = ?
vi = 22 m/s
θ = 40°
Solution:
(vi sin  ) 2
dy =
2g
(22m / s sin 40) 2
dy =
2(9.8m / s 2 )
dy = 10.20 m

41 Dorothy D. Silva 
 2. Given: g = -9.8.0 m/s2 dy = ?
vi = 30 m/s R=?
θ = 35°
Solution:
(vi sin  ) 2 (vi 2 sin 2 )
dy = R=
2g g
(30m / s sin 35) 2 (30 2 sin( 2 x35))
dy = R=
2(9.8m / s 2 ) 9.8m / s 2
dy = 15.11 m R = 86.30 m

3. Given: g = -9.8.0 m/s2 dy = ?

vi = 250 m/s t=?
θ = 37° R=?
Solution:
(vi sin  ) 2 2vi sin  (vi 2 sin 2 )
dy = t= R=
Examples: 2g g g
( 250m / s sin 37) 2 (2)(250m / s )(sin 37) ( 250 2 sin( 2 x37))
1. dy = on a football team
The punter t= it staysRin= the air for a long
2(9.8m / s 2 ) tries to kick a football
9.8mso/ sthat
2
9.8m / s 2 “hang
time”. If the ball is kicked with an initial velocity of 25.0 m/s at an angle of 60.0° above the
dy = 1154.91 m t = 30.70 s R = 6130.50 m
ground, what is the “hang time”?
4. Given: g = -9.8.0 m/s2 θ=?
2. When chasing a hare along a flat stretch of ground, a greyhound leaps into the air at a speed of
vi = 18 m/s
10.0 m/s, at an angle of 31° above the horizontal. (a) What is the range of his leap and (b) for how
R = 31 m
much time is he in the air?
Solution:
(vi 2 sin 2 )
3. A fire hose
R = ejects a stream of water at an angle of 35.0° above the horizontal. The water leaves
g
the nozzle with a speed of 25.0 m/s. Assuming that the water behaves like a projectile, how far
from a Θ
building
Rg the fire hose be located to hit the highest possible fire?
= ½ sinshould
-1

vi 2
4. The highest barrier (31 .8m / s 2 ) can clear is 13.5 m, when the projectile is lauched at an angle
m)(a9projectile
that
Θ = ½ sin -1

of 15.0° above the horizontal.

(18m / s What
)2 is the projectile’s launch speed?
Θ = 34.83°
5. A car drives straight off the edge of a cliff that is 54 m high. The police at the scene of the
accident note that the point of impact is 130 m from the base of the cliff. How fast was the car
traveling when it went over the cliff?

6. A motorcycle daredevil is attempting to jump across as many buses as possible. The takeoff ramp
makes an angle of 18.0° above the horizontal, and the landing ramp is identical to the takeoff ramp.
The buses are parked side by side, and each bus is 2.74 mwide. The cyclist leaves the ramp with a
speed of 33.5 m/s. What is the maximum number of buses over which the cyclist can jump?

7. A soccer player kicks the ball toward a goal that is 29.0 m in front of him. The ball leaves his foot
at a speed of 19.0 m/s and an angle of 32.0° above the ground. Find the speed of the ball when the
goalie catches it in front of the net.

8. An Olympic long jumper leaves the ground at an angle of 23° and travels through the air for a
horizontal distance of 8.7 m before landing. What is the takeoff speed of the jumper?

9. An airplane is flying with a velocity of 240 m/s at an angle of 30.0° with the horizontal. When the
42 from the plane. The flare hits the
altitude of the plane is 2.4 km, a flare is released Dorothy the 
targetD.onSilva
ground. What is the angle of the line of sight from the ground?
Solutions:
1. Given: vi = 22 m/s t=?
θ = 60°
Solution:
2vi sin 
t=
g
2(25m / s ) sin 60
t=
(9.8m / s 2 )
t = 4.42 s

2. Given: vi = 10 m/s R=?

θ = 31° t=?
Solution:

43 Dorothy D. Silva 
 (vi 2 sin 2 ) 2vi sin 
R= t=
g g
(10m / s) 2 sin( 2 x31) 2(10m / s ) sin 31
R= t=
9.8m / s 2 (9.8m / s 2 )
R = 9.01 m t = 1.05 s

3. Given: vi = 10 m/s R=?

θ = 31° t=?
Solution:
(vi 2 sin 2 ) R
R= R1/2 =
g 2
(25m / s ) 2 sin( 2 x35) 59.93m
R= R1/2 =
9.8m / s 2 2
R = 59.93 m R1/2 = 29.97 m

4. Given: dy = 13.5 m vi = ?
θ = 15°
Solution:
(vi sin  ) 2
dy =
2g
2gdy
vi =
sin 
2(9.8m / s 2 )(13.5m)
vi =
sin 15
vi = 62.85 m/s

5. Given: dy = 54 m vx = ?
dx = 130 m
Solution:
2dy
t= dx = (vx) (t)
g
2(54m) dx
t= vx =
9.8m / s 2 t
130m
t = 3.32 s vx = vx = 39.16 m/s
3.32 s

6. Given: vi = 33.5 m/s # of buses = ?

θ = 18°
Solution:

44 Dorothy D. Silva 
 (vi 2 sin 2 ) R
R= # of buses =
g 2.74m
(33.5m / s ) 2 sin( 2 x18) 67.31m
R= # of buses =
9.8m / s 2 2.74m
R = 67.31 m # of buses = 24

7. Given: R = 29 m vi = ?
vi = 19 m/s
θ = 32°
Solution:
(vi 2 sin 2 )
R=
g
Rg
vi =
sin 2
(29m)(9.8m / s 2 )
vi =
sin( 2 x32)
vi = 17.78 m/s

8. Given: R = 8.7 m vi = ?
θ = 23°
Solution:
(vi 2 sin 2 )
R=
g
Rg
vi =
sin 2
(8.7 m)(9.8m / s 2 )
vi =
sin( 2 x 23)
vi = 10.89 m/s

9. Given: dy = 2.4 km θ in the line of sight = ?

vi = 240 m/s
θ = 30°
Solution:
vi sin  2.4
t= R = (vix) (t) θ = tan-1
g 2.54
(240m / s ) sin 30
t= R = (240 m/s) (cos 30) (12.24s) θ = 43.38°
(9.8m / s 2 )
t = 12.24 s R = 25440.04 m = 2.54 m

45 Dorothy D. Silva 