Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 10.9 due 09/05/2021 at 11:59pm MST
Problem 1. (1 point)
Find the velocity, acceleration, and speed of a particle with posi-
tion function
r(t) = h−7tsin(t),−7tcos(t),7t2i
v(t) = h, , i
a(t) = h, , i
|v(t)|=
Solution:
SOLUTION
v(t) = r0(t) = h−(7sin(t) + 7tcos(t)),−(7 cos(t)−7tsin(t)),7·
2ti
a(t) = v0(t) = h7tsin(t)−14cos(t),14sin(t) + 7tcos(t),14i
|v(t)|=p(−(7sin(t) + 7tcos(t)))2+ (−(7cos(t)−7tsin(t)))2+ (7·2t)2=
√49 +245t2
Answer(s) submitted:
•-7sint-7tcost
•7tsint-7cost
•14t
•7tsint-14cost
•7tcost+14sint
•14
•
(score 0.857143)
Correct Answers:
•-[7*sin(t)+7*t*cos(t)]
•-[7*cos(t)-7*t*sin(t)]
•7*2*t
•-[7*cos(t)+7*cos(t)-7*t*sin(t)]
•7*sin(t)+7*sin(t)+7*t*cos(t)
•14
•sqrt(49+245*tˆ2)
Problem 2. (1 point)
Find the velocity and position vectors of a particle with accelera-
tion a(t) = 1k, and initial conditions v(0) = 3j+2kand r(0) =
3i−4j−4k.
v(t) = i+j+k
r(t) = i+j+k
Solution:
SOLUTION
a(t) = 1k⇒v(t) = R(1k)dt =1tk+Cand 3j+2k=v(0) = C,
so C=3j+2kand v(t) = 3j+ (2+1t)k
r(t) = R(3j+ (2+1t)k)dt =3ti+ (2t+ (1/2)t2)k+D.
But 3i−4j−4k=r(0) = D, so D=3i−4j−4kand
r(t) = 3i+ (3t−4)j+ (2t+ (1/2)t2−4)k
Answer(s) submitted:
•0
•3
•9t+2
•
•
•
(score 0.333333333333333)
Correct Answers:
•0
•3
•1*t + 2
•0*t + 3
•3*t + -4
•1*t*t/2 + 2*t + -4
1
Problem 3. (1 point)
Given that the acceleration vector is a(t)=(−cos(−t))i+
(−sin(−t))j+ (−3t)k, the initial velocity is v(0) = i+k, and
the initial position vector is r(0) = i+j+k, compute:
A. The velocity vector v(t) = i+j+k
B. The position vector r(t) = i+j+k
Solution:
SOLUTION
A.
v(t) = Ra(t)dt
= (sin(−t)))i+ (−cos(−t))j+−3
2t2k+C
and i+k=v(0) = −1j+C, so C=i+1j+kand
v(t) = (sin(−t) + 1)i+ (1−cos(−t))j+−3
2t2+1k
B.
r(t) = Rv(t)dt
= (cos(−t) + t)i+ (sin(−t) +t)j+−1
2t3+tk+D.
But i+j+k=r(0) = i+D, so D=j+kand
r(t) = (cos(−t) +t)i+ (sin(−t) + t+1)j+−1
2t3+t+1k
Answer(s) submitted:
•sin(-t)
•-cos(-t)
•
•
•
•
(incorrect)
Correct Answers:
•sin(-t)+1
•1-cos(-t)
•1-3*tˆ2/2
•cos(-t)+t
•sin(-t)+t+1
•t-3*tˆ3/6+1
Problem 4. (1 point)
The position function of a particle is given by r(t) =
h−2t2,−3t,t2−8ti.
At what time is the speed minimum?
Solution:
SOLUTION
r(t) = h−2t2,−3t,t2−8ti ⇒ v(t) = h−4t,−3,2t−8i
Speed = |v(t)|=p16t2+9+ (2t−8)2=√20t2−32t+73 .
To detemine when the speed is minimum, we first compute the
derivative of the speed:
d
dt |v(t)|=1
2(20t2−32t+73)−1/2(40t−32).
This is zero if and only if the numerator is zero, that is,
40t−32 =0 or t=4
5.
Since d
dt |v(t)|<0 for t<4
5and d
dt |v(t)|>0 for t>4
5, the mini-
mum speed is attained at t=4
5.
Answer(s) submitted:
•.3
(incorrect)
Correct Answers:
•0.8
Problem 5. (1 point)
A dense particle with mass 8 kg follows the path r(t) =
hsin(2t),cos(5t),2t11/2iwith units in meters and seconds.
What force acts on the mass at t=0?
h, , ikg m/s2
Solution:
SOLUTION
r(t) = hsin(2t),cos(5t),2t11/2i
⇒v(t) = r0(t) = h2cos(2t),−5sin(5t),11t
9
2i
⇒a(t) = v0(t) = −4sin(2t),−25cos(5t),99
2t
7
2
⇒a(0) = h0,−25,0i
By Newton’s Second Law, at t=0,
F(0) = ma(0) = h0,8(−25),0i=h0,−200,0iis the required
force.
Answer(s) submitted:
•0
•-15
•0
(score 0.666667)
Correct Answers:
•0
•-200
•0
2
Problem 6. (1 point)
A projectile is fired from ground level with an initial speed of 150
m/sec and an angle of elevation of 30 degrees. Use that the accel-
eration due to gravity is 9.8 m/sec2.
(a) The range of the projectile is meters.
(b) The maximum height of the projectile is me-
ters.
(c) The speed with which the projectile hits the ground is
m/sec.
Solution:
SOLUTION
We set up the axes so that the projectile starts at the origin. Then
r(0) = 0.
|v(0)|=150 and, since the angle of elevation is 30o,
v(0) = 150cos(300)i+150 sin(30o)j=75√3i+75j.
Ignoring air resistance, the only force is that due to gravity, so
a(t) = −9.8jand, integrating, we have
v(t) = −9.8tj+C.
But 75√3i+75j=v(0) = C, so v(t) = (75√3)i+ (75 −9.8t)j.
Integrating again gives r(t)=(75√3t)i+(75t−4.9t2)j+Dwhere
0=r(0) = D.
Thus the position function of the projectile is
r(t) = (75√3t)i+ (75t−4.9t2)j
(a) Parametric equations for the projectile are
x(t) = 75√3t,y(t) = 75t−4.9t2.
The projectile reaches the ground when y(t) = 0 ( and t>0)
⇒75t−4.9t2=t(75 −4.9t) = 0⇒t=75
4.9.
So the range is x75
4.9=75√375
4.9≈1988.32.
(b) The maximum height is reached when y(t)has a critical num-
ber (or, equivalently, when the vertical component of the velocity
is 0): 75 −9.8t=0⇒t=75
9.8.
Thus the maximum height is y75
9.8=7575
9.8−4.975
9.82≈
286.99
(c) From part (a), impact occurs at t=75
4.9. Thus, the velocity at
impact is v75
4.9= (75√3)i+ (75 −9.8(75
4.9)j= (75√3)i−75j
and the speed is
p3(75)2+ (75)2=150
Answer(s) submitted:
•
•
•150
(score 0.333333333333333)
Correct Answers:
•1988.32363113774
•286.989795918367
•150
Problem 7. (1 point)
A ball is thrown at an angle of 45 degrees to the ground, and lands
90 meters away.
The initial speed of the ball was m/sec.
Solution:
SOLUTION
Let v0be the initial speed. We set up the axes so that the projectile
starts at the origin. Then r(0) = 0.
Since the angle of elevation is 45o,
v(0) = v0cos(450)i+v0sin(45o)j=v0
√2
2i+v0
√2
2j.
Ignoring air resistance, the only force is that due to gravity, so
a(t) = −9.8jand, integrating, we have
v(t) = −9.8tj+C.
But v0
√2
2i+v0
√2
2j=v(0) = C, so
v(t) = v0
√2
2!i+ v0
√2
2−9.8t!j.
Integrating again gives
r(t) = v0
√2
2t!i+ v0
√2
2t−4.9t2!j+Dwhere 0=r(0) = D.
Thus the position function of the projectile is
r(t) = v0
√2
2t!i+ v0
√2
2t−4.9t2!j
Parametric equations for the ball are
x(t) = v0
√2
2t,y(t) = v0
√2
2t−4.9t2.
The ball lands when y(t) = 0 ( and t>0) ⇒v0
√2
2t−4.9t2=
t v0
√2
2−4.9t!=0⇒t=v0√2
9.8.
Now, since it lands 90 meters away, 90 =x v0√2
9.8!=
v0
√2
2
v0√2
9.8=v2
0
9.8, and the initial velocity is
v0=p9.8(90) = √882.
Answer(s) submitted:
•29.69
(correct)
Correct Answers:
•29.698484809835
3
Problem 8. (1 point)
A body of mass 4 kg moves in a (counterclockwise) circular path
of radius 6 meters, making one revolution every 5 seconds. You
may assume the circle is in the xy-plane, and so you may ignore
the third component.
A. Compute the centripetal force acting on the body.
h,i
B. Compute the magnitude of that force.
Note: Use exact forms or at least 4 significant digits in your
answers.
Solution:
SOLUTION:
Since the body makes one revolution every 5 seconds, the angular
frequency is 2
5π.
The position function of the body is then
r(t) = 6 cos2
5πt,6sin2
5πt.
The acceleration is then
a(t) = r00(t) = −24
25 π2cos2
5πt,−24
25 π2sin2
5πt.
By Newton’s Second Law,
F=ma(t) = −96
25 π2cos2
5πt,−96
25 π2sin2
5πt.
The magnitude of the force is 96
25 π2.
Answer(s) submitted:
•
•
•
(incorrect)
Correct Answers:
•- 4 * 7.5398223686155**2 / 6 * cos(7.5398223686155 * t / 6)
•- 4 * 7.5398223686155**2 / 6 * sin(7.5398223686155 * t / 6)
•37.8992809001831
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