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 θ 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 +

vji

64 +

?6424.312.6

=+=+=

vjiajiv av

( )

jivv

jivji

tva

t

tav

36.32.602.0

24.312.664

67.3

67.3

+==

+−=+⇒

∆∆

=

=

=

v

a

θ

( ) ( )0

2222

852785509644

64363266436326

...cos

cos..jij.i.

cosavav

=⇒=

+

+=+⋅+

=⋅

θθ

θ

θ

2. The y-coordinate of a particle in curvilinear motion is given by y = 4t3−3t, 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

2204

+=⇒=−

=⇒=⇒= ∫∫

tvtv

tdtdvdtadvdt

dva

xx

tv

xxxx

xx

aya

xaxvθ

yvv

αo

x

y

yx

.vv

tana

s/m.vvv

9841

451322

=

=

=+=

θ o

x

y

yx

.aa

tana

s/m.aaa

4363

8326 222

=

=

=+=

α

3. 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

32

320

22

2

40

4

330

2

2

2

2000

txtx

dttdxtdtdxv

tv

dttvdtadv

tdt

dva

txx

x

tx

tx

v

x

xx

x

=⇒=−

=⇒==

=

=−⇒=

==

∫∫

∫∫∫

( )

( )32

233

2202

2

2121

281

322

21

32

221

2442

8

88

−=

−=

−=

−=

+=⇒=−

=⇒=

==⇒==

∫∫

yx

yyx

yt

tyty

tdtdydtdyv

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.

s/mvvv

s/mvtvs/mvtv

sttxmx

yx

yy

xx

302418

248182

3183

218

2222

2

3

=+=+=

=⇒==⇒=

=⇒==⇒=

4. 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.

5. 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=−g ax=0

( ) 22 819212173020

21 t.t.sinRgttvyy yoo −+=⇒−+= (2) at B

(1) ⇒ t=0.0094R

(2) ⇒ ( ) ( )

m.R

R..R..sinR

22967

009408192100940217320 2

=

−=

6 . 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 ( )( ) θ

θsinv

cosv

yo

xo

40

40

−=

=

Horizontal: ( )θ

θcos

ttcostvxx xoo 214020 =⇒=⇒+=

Vertical:

ay=−g ax=0

( ) ( ) 22 8192140081

21 t.tsin.gttvyy yoo −−+=−⇒−+= θ

( )( )( )

( )

5866381

3551602860

226125740226142020

05740202261

0812012261

1226120812

181921

214081

2

212

2

2222

..

..

x

...

x.xx.

xtan.tantan.

tansecsec.tan.cos

.cos

sin.

,

−→

−→

−−±−=⇒=−+

==−++

+=−−=−⇒

−−=−

θ

θθθ

θθθθθθ

θ

20 m

d 1.638o d=20 tan1.638=0.572 m h=2.8−(1+0.572)=1.228 m

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

and θ=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

x θ

y=−0.05x2

y

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

x θ=60ο

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=−g ax=0

( )2

2

9054391021

t.t.y

gttvyy yoo

−=

−+=

( )

s.ttt.t.t.t.t.

t.t.t.x.y

35301053391008190543910

605090543910050

2

1222

222

==⇒−⇒=+−

−=−⇒−=

s/m.vvv yx 262322 =+=

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 and of pin P when x = 60 mm. v a

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

C x

y

10. A projectile is launched with speed v0 from point A. Determine the launch angle θ which results in the maximum range R up the incline of angle α (where 0≤ α ≤ 90°). Evaluate your results for α = 0, 30° and 45°.