a particle moves in the plane with a y-component of velocity...
TRANSCRIPT
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°.