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Ch 4.2
#8
A particle initially located at the origin has an acceleration of a = 3 j m/s2 and an initial velocity
of v = 500 i m/s. Find (a) the vector position and velocity at any time t and (b) the coordinates
and speed of the particle at t = 2.00 seconds.
(a)
r(t) = (½ 3j t2 + 500i t) meters
(b)
v(t) = (3j t + 500i ) m/s
Ch 4.3
r(2) = (6 j + 1000 i ) meters
or (1000, 6)
v(t) = (3j t + 500i ) meters
#19
A place kicker must kick a football from a point 36 m from the goal. Half the crowd hopes the
ball will clear the crossbar, which is 3.05 meters high. When kicked, the ball leaves the ground
with a speed of 20 m/s and an angle of 53 degrees above the horizontal.
(a) By how much does the ball clear or fall short of the crossbar?
(b) Does the ball approach the crossbar while still rising or while falling?
x = 36 m
y = 3.05 m
 = 53(a 3-4-5 )
v = 20 m/s
vy = sin 53 20 m
vy = 16 m/s
(a)
dy = ½ at2 + vy-init t + dy-init
dy = ½ -9.8(3)2 + 16 (3)
dy = 3.9 meters @ 3 seconds
vx = cos 53 20 m
vx = 12 m/s
Ch 4.4
vx = x / t
12 m/s = 36 / t
t = 3 sec
3.9 m clears the 3.05 m bar by
0.85 meters
(b) dy = ½ at2 + vy-init t + dy-init
3.05 = -4.9t2 + 12 t + 0
4.9t2 - 16 t + 3.05 = 0
(16  (144 – 4(4.9)3.05)½) / 2(4.9)
t = (16  14) / 9.8
t = 3.06 sec, 0.20 sec
12 m/s = x / 3.06 seconds
x = 36.72 m
It barely makes it over the bar;
thus descending.
#32
The astronaut orbiting the Earth in Figure P4.32 is preparing to dock with a Westar VI
satellite. The satellite is in a circular orbit 600 km above the Earth’s surface, where the freefall acceleration is 8.21 m/ s2. The radius of the Earth is 6,400 km. Determine the speed of
the satellite and the time required to complete one orbit around the earth.
a
=
v2
/
r
t = 5800 seconds
6 2
6 2
8.21 = (2*(6.4 + .6)*10 )
/ (6.4 + .6)*10 t
t = 1 hour & 36 2/3 minutes
Ch 4.5
#35
The figure below represents
the total acceleration of a
particle moving clockwise in a
circle of radius 2.50 meters at
a certain instant of time. At
this instant, find
(a) the radial acceleration
(b) the speed of the particle
(c) its tangential acceleration
Ch 4.6
(a)
ar = cos 30 a
ar = 13 m/s2
(b) ar = v2 / r
13m/s2 = v2 / 2.5 m
v
= 5.7 m/s
(c) a2 = ar2 + at2
152
= 132 + at2
at
= 7.5 m/s2
#38
Heather in her Corvette accelerates at the rate of (3.00 i – 2.00 j) m/s2, while Jill in her
Jaguar accelerates at (1.00 i + 3.00 j) m/s2. The both start from rest at the origin of the xy
coordinate system. After 5.00 s, (a) what is Heather’s speed with respect to Jill, (b) how far
apart are they, and (c) what is Heather’s acceleration relative to Jill?
Integration tutorial 
Other hints
v = ∫ a dt
 opposite to (a = dv/dt)
x = ∫ v dt
 opposite to (v = dx/dt)
∫dt
∫ t0 dt
 raise the index by one
 take the new index and place it in
the denominator
1
t / 1 (from ti to tf)
HEATHER v(t) = (3∫dt i – 2∫dt j)
v(t) = (3t i – 2t j) m/s
v(5) = (15 i – 10 j) m/s
x(t) = (3t∫dt i – 2t∫dt j)
x(t) = (1.5 t2 i – t2 j) m/s
x(5) = (37.5 i – 25 j) m/s
v(t) = (1∫dt i + 3∫dt j)
JILL
v(t) = (t i + 3t j) m/s
v(5) = (5 i + 15 j) m/s
x(t) = (t∫dt i + 3t∫dt j)
x(t) = (½ t2 i + 1.5t2 j) m/s
x(5) = (12.5 i+ 37.5 j) m/s
Ch 4
Try it for
∫ t dt
∫ t1 dt
½ t2 (from ti to tf)
(a) |v(5)H| -|v(5)J| = ((15-5) i + (-1015) j)
|v(5)H| -|v(5)J| = ((10) i + (-25) j) m/s
|v(5)H| -|v(5)J| = (102 + -252)1/2
|v(5)H| -|v(5)J| = 26.9 m/s
(b) |x(5)H| -|x(5)J| = ((37.5-12.5)i + (-2537.5)j)
a = dv/dt
a ∫dt = ∫dv
v = dx/dt
v ∫dt = ∫dx
|x(5)H|-|x(5)J| = ((25)i + (-62.5)j) m
speed = |v|
|x(5)H|-|x(5)J| = (252 + -62.52)1/2
|x(5)H|-|x(5)J| = 67.3 m
(c)
aH - aJ = (3.00 i – 2.00 j) - (1.00 i + 3.00 j)
aH - aJ = (2 i – 5 j) m/s2
#63
A car is parked on a steep incline overlooking the ocean, where the
incline makes an angle of 37.0° below the horizontal. The negligent
driver leaves the car in neutral, and the parking brakes are
defective. Starting from rest at t = 0, the car rolls down the
incline with a constant acceleration of 4 m/s2, traveling 50 m to the
edge of the vertical cliff. The cliff is 30 m above the ocean. Find
(a) the speed of the car when it reaches the edge of the cliff and
the time at which it arrives there, (b) the velocity of the
car when it lands in the ocean, (c) the total time interval that the car is in motion, and (d) the
position of the car when it lands in the ocean, relative to the base of the cliff.
Given:
a = 4 m/s2
r
= ½at2 + vot + ro
rincline = (½(-4)t2 + 50) m
v = dr / dt
v = -4t m/s
(a) Use point of view perspective where
my zero point is the top and it’s speeding
up thus (+) acceleration
rincline
= ½ a t2
2
50
=½4t
tincline = 5.00 sec
v=4t
v = 4 (5)
v = 20 m/s
 from above
(b)
vy = v cosθ
vy = 20 cos53
vy = 12.0 m/s
vx = v sinθ
vx = 20 sin53
vx = 16.0 m/s
a = Δv / Δt
-10 = (vy-f – -12) / 1.53 sec
vy-f = - 27.3 m/s
v = (16 i - 27.3) j ) m/s
(c) dy = ½ gt2 + vy-i t + dy-i
0 = ½-10t2 + -12 t + 30
t2 + 2.4 t = 6
(t + 1.2)2 = 6 + 1.22
tcliff = 1.53 seconds
ttotal = 5.00 + 1.53 = 6.53 s
(d)
dx = 16 (1.53)
dx = 24.5 meters
or
r = 24.5 i meters
Ch 4.1
#2
A golf ball is hit off a tee at the edge of a cliff. Its x & y coordinates as functions of time are
given by the following expressions:
x = 18 t
or with units
x = 18m/s * t
2
y = 4 t – 4.9 t
or with units
y = 4m/s * t – 4.9m/s2 * t2
(a) Write a vector expression for the ball
position as a function of time, using the unit
vectors i and j.
r(t) = (18t) i + (4t – 4.9t2) j
or
r(t) = (18t) i + (4t – ½ g t2) j
By taking the derivatives, obtain
expressions for (b) the velocity vector, v,
as a function of time.
v = dr/dt = 18 i + (4 – 9.8 t) j
(c) the acceleration vector, a, as a function
of time.
a = dv/dt = (0) i + (-9.8) j
a = dv/dt = -(9.8 m/s2) j (w/units)
Ch 4.2
Next use unit-vector notation to write expressions
for
(d) the position
r(t) = (18t) i + (4t – 4.9t2) j
r(3) = (18*3) i + (4*3 – 4.9*32) j = 54 i – 32.1 j
r(3) = 54 i – 32.1 j
(e) the velocity
v(3) = (18) i + (4 – 9.8 *3) j
v
= (18) i - (25.4) j
(f) the acceleration of the golf ball
all at time t = 3.00 seconds.
a = dv / dt
a = -(9.8 m/s2) j (w/units)
#5
At t = 0, a particle moving in the xy plane with a constant acceleration has a velocity of v i = 3i –
2j m/s and is at the origin. At t = 3 seconds, the particles velocity is v = (9i + 7j) m/s. Find
(a) the acceleration of the particle
(b) its coordinates at any time t.
a = v / t
r = ro + vot + ½ at2
a = (9 – 3)i + (7 - -2)j / 3
r(t) = 0 + (3i – 2j)*t + ½ (2i +3j)*t2
a = 6i + 9j / 3
a = 2i +3j m/s2
x(t) = 3t + t2
y(t) = -2t + 3t2/2
Ch 4.3
#22
A dive bomber has a velocity of 280 m/s at an
angle  below the horizontal. When the
altitude of the aircraft is 2.15 km, it releases
a bomb, which hits a target on the ground.
The magnitude of the displacement from the
point for release of the bomb to the target is
3.25 km. Find the angle, .
x
= ½ at2 + vx-init t + xo
2437 = 0 + 280 cos t + 0
t = 8.7 / cos
y
= ½ at2 + vy-init t + yo
2150 = ½ 10t2 + 280 sin t
430 = t2 + 56 sin t
where t = 8.7 / cos
430 = 75.8 / cos2 + 487 tan
Plug in excel and increment  until left side equals right side.
0.583 radians or 33.4
xf 
 3.25 km  2   2.15 km  2
y f  x f tan  
 2.437 km
gx2f
2vi2 cos2  i
 9.8 m s   2 437 m 
2150 m   2 437 m  tan  
2
i
2
2 280 m s cos2  i
2
 2150 m   2 437 m  tan  i   371.19 m
  1 tan2  i
 tan 2   6.565tan  i  4.792  0
 tan  i 
Ch 4.4

1
6.565 
2
 6.565 2  4 1 4.792

 3.283  3.945.
#31
Young David who slew Goliath experimented with slings before tackling the giant. He found that
he could revolve a sling of length 0.6 m at the rated of 8 rps. If he increased the length to 0.9
m, he could revolve the sling only 6 rps. (a) Which rate of rotation gives the greater speed for
the stone at the end of the sling? (b)What is the centripetal acceleration of the stone at 8
rps? (c) What is the centripetal acceleration at 6 rps?
(a)
a = v / t = v2 / r
v = 8 rev / s * 2r.6 / rev
a = (*2r / t)2 / r
v = 16*0.6
a = (2)22r / t2
v = 9.6 m/s
(b)
a = (2)22r / t2
v = 6 rev / s * 2r.9 / rev
a = (2*8)22(0.6) / 12
v = 12*0.9
a = 1516 m/s2
v = 10.8 m/s
(c)
a = (2)22r / t2
Thus at 6 rps results in a greater tangential
a = (2*6)22(0.9) / 12
velocity
a = 1279 m/s2
Ch 4.6
#41
A river has a steady speed of ½ m/s. A student swims upstream a distance of 1 km and swims
back to the starting point. If the student can swim at a speed of 1.2 m/s in still water how long
does the trip take?
Compare this with the time the trip would take if the water were still.
(a)
vnet up = 1.2 m/s – 0.5 m/s
ttotal = tnet up + tnet down
vnet up = 0.7 m/s
ttotal = 1428.6 s + 588.2 s
tnet up = 1000 m / 0.7 m/s
ttotal = 2017 s
tnet up = 1428.6 sec
vnet down = 1.2 m/s + 0.5 m/s
vnet down = 1.7 m/s
tnet down = 1000 m / 1.7 m/s
tnet down = 588.2 sec
Ch 4
(b)
total time w/o current = 2000 m / 1.2 m/s
total time w/o current = 1667 seconds
(2017 – 1667) / 1667 = 21.0%
w/ current takes 21 % longer than w/o current
#56
A boy can throw a ball a maximum horizontal distance
of 40 m on a level playing field. How far can he throw
the same ball vertically upward?
Assume that his muscles give the ball the same speed in each case.
vx = cos  v vy = sin  v
a = v / t
-10 = (0 - sin v) / ttop
ttop = v sin / 10
ttotal = ttop + tbottom
Level Ground: ttop = tbottom
ttotal = v sin / 5
vx
= x / ttotal
vcos = R / (v sin / 5)
v2
= 5R / sincos
v
= (10r)½ m/s
Thrown the ball straight UP
…the only force on it is gravity
a
= v
/ t
½
-10
= (0 - (10r) ) / t
t
= (r/10)½ seconds
dy
dy
dy
= ½ at2
= ½ -10(r/10)½*2
=½r
+
vot
+ (10r)½* (r/10)½
Let’s solve again but this time the total distance traveled is r instead of 40 meters.
vx = cos  v
vy = sin  v
ay = v / t
-10 = (0 - sin  v) / t
solve for time
t = sin  * v / 10
but this is only half of the time. Thus the ttotal = (2 * sin  v / 10) = sin  * v / 5
We also know the vx
= dx / ttotal
where vx = cos  * v
cos  v
=r
/ sin  v / 5
2
v = 5 r / (sin  cos )
Where  must be 45 since we know it was thrown at the maximum
distance.
v2 = 5 r / (0.707 * 0.707) = 10 r
v = (10r)½
So now we know the maximum velocity the ball can be thrown, which is now pointed up.
-10 = (0 – (10r)½) / t
t = (r/10)½ seconds
dy = ½ at2 + vyot = ½ -10(r/10)½*2 + (10r)½* (r/10)½ = ½ r
Ch 4
#57
A stone at the end of a sling is whirled in a vertical circle of radius 1.2 m at a constant speed v i
= 1.5 m/s as in Figure P4.57. The center of the string is 1.5 m above the ground. What is the
range of the stone if it is released when the sling is inclined at 30.0 with the horizontal (a) at
A? (b) at B? What is the acceleration of the stone (c) just before it is released at A? (d) just
after it is released at A?
vy = sin 60*1.5
vx = cos 60*1.5
vy = 1.3 m/s
vx = 0.75 m/s
(a)
ay
= vy / t
-9.8 = (0-1.3) / t
ttop
= 0.133 sec
dy-init = 0.3+1.2+sin30*1.2
dy-init = 2.1 m
dtop = ½ at2 + vot + dy-init
dtop = 0.75 t – 5 t2 + 2.1
dtop = 2.185 m
dbottom = ½ at2 + vy-init + dy-ini
2.185 = ½ 9.8 t2
tbottom = 0.668 sec
ttotal = ttop + tbottom
ttotal = 0.133 + 0.668
ttotal = 0.8 sec
vx = x / t
0.75 m/s = x / 0.8 sec
x = 0.60 meters
dabove ground = vo in y compt + ½ at2
2.1 m
=
1.2 t + 5t2
5t2 + 1.2t – 2.1 = 0
[-b  (b2 – 4ac)1/2] / 2a
(-1.2  6.6) / 10
t = 0.54 seconds
vx = x / t
0.75 m/s = x / 0.54 sec
x = 0.40 meters
(b)
(c)
a = v / t
a = v2 / r
a = 2.25 / 1.2
a= 1.875 m/s2
Ch 4
(d)
immediately after it’s release ANYWHERE, the
only force acting on it (neglecting air friction) is
gravity which has a rate of acceleration of 9.8
m/s2 pointed down
#65
The determined coyote is out once more to try to capture the elusive roadrunner. The coyote
wears a pair of Acme jet-powered roller skates, which provide a constant horizontal
acceleration of 15 m/s2. The coyote starts off at rest 70 m from the edge of a cliff at the
instant the roadrunner zips past him in the direction of the cliff. (a) if the roadrunner moves
with a constant speed, determine the minimum speed he must have to reach the cliff before the
coyote. At the brink of the cliff, the roadrunner escapes by making a sudden turn, while the
coyote continues straight ahead. (b) If the cliff is 100 m above the floor of a canyon,
determine where the coyote lands in the canyon (assume his skates remain horizontal and
continue to operate when he is in flight). (c) Determine the components of the coyote’s impact
velocity.
(a)
d = do + vot + ½ at2
Coyote distance = ½ at2
70 m = ½ 15 t2
t = 3.055 seconds
Thus the road runner must also cover the same distance at a constant velocity in 3.055 seconds.
vave = d / t
vroad runner = 70 m / 3.055 s = 22.9 m/s
(b) The canyon is 100 m deep, thus the coyote is in free fall for 100 m
100 m = do y + vo yt + ½ ayt2 = ½ 10t2 = 4.47 seconds
(c) d = do + vot + ½ at2
dx = ½ at2 = ½ 15 (7.52)2 = 424 meters total in the x-direction
70 m was prior to the coyote leaving the cliff, so the coyote landed
canyon.
354 meters into the
(d) Total time the coyote was accelerating in the x direction is 3.055 + 4.47 = 7.52 seconds
a = v / t = 15 = v / 7.52 seconds
vf  x dir = 113 m/s
10 = v / 4.47
vf  y dir = 44.7 m/s