1. The velocity of a particle moving in the x-y plane is given by m/s

at time t=3.65 s. Its average acceleration during the next 0.02 s is m/s2.

Determine the velocity of the particle at t=3.67 s and the angle q between the

average-acceleration vector and the velocity vector at t=3.67 s.

Solution:

at t=3.65 s, (during 0.02 s)

at t=3.67 s,

ji

24.312.6

v

ji

64

?

6424.312.6

v

jiajiv av

jivv

jivji

t

va

t

tav

36.32.6

02.0

24.312.664

67.3

67.3

v

a

q

0

2222

85278550

9644

64363266436326

..

.cos

cos..jij.i.

cosavav

qq

q

q

2. The position of a particle is defined by Ԧ𝑟 = 5𝑐𝑜𝑠2𝑡Ԧ𝑖 + 4𝑠𝑖𝑛2𝑡Ԧ𝑗, [m], where t is

in seconds and the arguments for the sine and cosine are given in radians.

Determine the magnitudes of the velocity and acceleration of the particle when

t=1 s. Also prove that the path of the particle is elliptical.

3. The y-coordinate of a particle in curvilinear motion is given by y = 4t33t, where y is in meters

and t is in seconds. Also, the particle has an acceleration in the x-direction given by ax = 12t m/s2.

If the velocity of the particle in the x-direction is 4 m/s when t = 0, calculate the magnitudes of

the velocity and acceleration of the particle when t = 1 s. Construct and in your solution.

Solution:

when t=1 s vy = 9 m/s , ay = 24 m/s2

ax=12t

when t=1 s vx = 10 m/s , ax = 12 m/s2

v

a

v

a

taytvytty yy 2431234 23

4664

12

22

04

tvtv

tdtdvdtadvdt

dva

xx

tv

xxxx

x

x

a

ya

xaxvq

yv

v

o

x

y

yx

.v

vtana

s/m.vvv

9841

451322

q

o

x

y

yx

.a

atana

s/m.aaa

4363

8326 222

4. A particle moves in the x-y plane with a y-component of velocity in meters/second given

by vy=8t with t in seconds. The acceleration of the particle in the x-direction in meters per

second squared is given by ax=4t with t in seconds. When t=0, y=2 m, x=0 and vx=0. Find

the equation of the path of the particle and calculate the magnitude of the velocity of the

particle for the instant when its x-coordinate reaches 18 m.

Solution:

x direction y direction

3

2

3

20

22

2

40

4

33

0

2

0

2

2

000

tx

tx

dttdxtdt

dxv

tv

dttvdtadv

tdt

dva

tx

x

x

t

x

t

x

v

x

xx

x

32

233

22

02

2

212

1

28

1

3

22

2

1

3

2

22

1

2442

8

88

yx

yyx

yt

tyty

tdtdydt

dyv

s/madt

dvtvy

/

ty

y

yy

y

Equation of the path

calculate the magnitude of the velocity of the particle for the instant when its x-

coordinate reaches 18 m.

smvvv

smvtv

smvtv

stt

xmx

yx

yy

xx

/302418

/248

/182

3183

218

2222

2

3

5. The skateboard rider leaves the ramp at A with an initial velocity vA at a 30o angle. If

he strikes the ground at B, determine vA and the time of flight.

6. The boy at A attempts to throw a ball over the roof of a barn such that it is launched at

an angle qA=40o. Determine the minimum speed vA at which he must throw the ball so

that it reaches its maximum height at C. Also, find the distance d where the boy must

stand so that he can make the throw.

7. Projectile (1) is fired with a speed of v=60 m/s at an angle of 60o. Projectile (2) is then

fired with the same speed 0.5 s later. Determine the angle q of the second projectile so

that the two projectiles collide. At what position (x,y) will this happen?

8. For a certain interval of motion, the pin P is forced to move in the fixed parabolic

slot by the vertical slotted guide, which moves in the x direction at the constant rate

of 40 mm/s. All measurements are in mm and s. Calculate the magnitudes of Ԧ𝑣 and

Ԧ𝑎 of pin P when x = 60 mm.

SOLUTION

Trajectory of the pin is160

2xy ??60 avmmxwhen

jaiaajvivv yxyx

jaiaajvivv yxyx

yaxayvxv yxyx ,,

0)(/40 xacstsmmxv xx

The first derivative of trajectory equation with respect to time

smm

xxxxy

xy /30

80

4060

80160

2

160

2

jiv

3040 Magnitude of velocity of the pin: smmv /503040 22

22

2 /2080

40

80

1

80smmxxxy

xx

dt

dy

dt

d

ja

20

Magnitude of acceleration of the pin:

2/20 smma

9. Pins A and B must always remain in the vertical slot of yoke C, which

moves to the right at a constant speed of 6 cm/s. Furthermore, the pins

cannot leave the elliptic slot. What is the speed at which the pins approach

each other when the yoke slot is at x = 50 cm? What is the rate of change

of speed toward each other when the yoke slot is again at x = 50 cm?

100 cm

60 cm

x

6 cm/s

yoke

Cx

y

100 cm

60 cm

x

6 cm/s

yoke

C x

y

Equation of elliptical slot is 160100 2

2

2

2

yx

??50 avcmxwhen

jaiaajvivv yxyx

cmycmxfor 96.5150

Yoke C moves at a constant speed, so 0/6 xvacstscmxv xxx

Using the equation of trajectory 160100 2

2

2

2

yx

First derivative of the trajectorywith respect to time

0

60

96.51

100

6500

60

2

100

22222

yyyxx

scmyvy /078.2 Velocity of pin B jivB

078.26

Velocity

(For pin A )scmvcmy y /078.296.51

Acceleration

Second derivative of the trajectory equation with respect to time

0

60

96.51

60

078.2

100

6

06060100100

22

2

2

2

2222

y

yyyyxxxx

2/3325.0 scmy

ja

3325.0 (For pin B)

10. A long-range artillery rifle at A is aimed at an angle of 45o with the

horizontal, and its shell is just able to clear the mountain peak at the top of its

trajectory. Determine the magnitude u of the muzzle velocity, the height H of the

mountain above sea level, and the range R to the sea.

x

y

Horizontal:

Vertical:

SOLUTION

tvxxuvvaxooxoxx 45cos0

2

2

145sin gttvyygtugtvvga

yooyoyy

B

C

Point B (at apex) BByoy tutgugtvv 873.13045sin0

81.9

1

B

Bxoot

ututvxx11313.7

45cos08000

21 = smustB /18.39655.28

22 55.2881.92

155.2845cos18.3960

2

1 Byoo ygttvyy

myHmy BB 46006004000

Point C (sea level)

0600142809054

8192

145183960600

2

1

2

2

2

CC

CC

yoo

t.t.

t.tsin.

gttvyy

06.2

18.59

Ct

m.R

.cos.R

tvxxxoo

7816578

185945183960

2

11. A projectile is launched with an initial speed of 200 m/s at an angle

of 60o with respect to the horizontal. Compute the range R as measured

up the incline.

x

y

s/m.sinv

s/mcosv

yo

xo

217360200

10060200

Horizontal:

tcosRtvxx xoo 100020 (1) at B

Vertical:

ay=gax=0

22 8192

12173020

2

1t.t.sinRgttvyy yoo (2) at B

(1) t=0.0094R

(2)

m.R

R..R..sinR

22967

009408192

100940217320

2

SOLUTION

12 . Determine the location h of the spot which the pitcher must throw

if the ball is to hit the catcher’s mitt. The ball is released with a speed of

40 m/s.

x

y

q

q

sinv

cosv

yo

xo

40

40

Horizontal: q

qcos

ttcostvxx xoo2

14020

Vertical:

ay=gax=0

22 8192

140081

2

1t.tsin.gttvyy yoo q

586

6381

35516

02860

22612

574022614202005740202261

0812012261

1226120812

1819

2

1

2

14081

2

212

2

2222

.

.

.

.x

.

..x.xx.

xtan.tantan.

tansecsec.tan.cos

.cos

sin.

,

q

qqq

qqqqqq

q

20 m

d1.638o

d=20 tan1.638=0.572 mh=2.8(1+0.572)=1.228 m

d

SOLUTION

13. A golfer strikes the ball at the top of a hill with an initial velocity of vo=12 m/s

and q=60o with the horizontal. Determine the coordinates of the point the ball hits

the ground, its velocity at this moment and its total flight time. The equation of

curvature of the hill is given by y= 0.05x2. Dimensions are given in meters. Take

the gravitational attraction of the earth as g=9.81 m/s2.

vo

xq

y=0.05x2

y

B

Determine the coordinates of the point the ball hits the ground, its velocity at this moment and its total

flight time.

vo=12 m/s

xq60o

y=0.05x2

y

O

B

For B

For O

(t2 is the total flight time.)

Coordinates of B: xB=20.1 m , yB=20.24 m

Impact velocity: vx=6 m/s (cst), vy=(vo)y gt=10.39 9.81(3.35)=22.47 m/s

s/m.sinv

s/mcosv

yo

xo

39106012

66012

Horizontal: ttvxx xoo 6

Vertical:

ay=gax=0

2

2

90543910

2

1

t.t.y

gttvyy yoo

st

tttttt

tttxy

35.3

00105.339.1008.1905.439.10

605.0905.439.1005.0

2

1

222

222

s/m.vvv yx 262322