Austin Cholley Zhu MAT 267 ONLINE A Spring 2021
Assignment Section 10.9 due 01/28/2021 at 11:59pm MST
1. (1 point) Find the velocity, acceleration, and speed of a
particle with position function
r(t) = htsin(t),tcos(t),−t2i
v(t) = h, , i
a(t) = h, , i
|v(t)|=
Solution:
SOLUTION
v(t) = r0(t) = hsin(t) + tcos(t),cos(t)−tsin(t),−(2t)i
a(t) = v0(t) = h2cos(t)−tsin(t),−(2sin(t) +tcos(t)),−2i
|v(t)|=p(sin(t) +tcos(t))2+ (cos(t)−tsin(t))2+ (−(2t))2=
√1+5t2
Answer(s) submitted:
•tcos(t)+sin(t)
•-tsin(t)+cos(t)
•-2t
•-tsin(t)+2cos(t)
•-tcos(t)-2sin(t)
•-2
•sqrt((tcos(t)+sin(t))ˆ(2)+(-tsin(t)+cos(t))ˆ(2)+(-2t)ˆ(2))
(correct)
Correct Answers:
•sin(t)+t*cos(t)
•cos(t)-t*sin(t)
•-(2*t)
•cos(t)+cos(t)-t*sin(t)
•-[sin(t)+sin(t)+t*cos(t)]
•-2
•sqrt(1+5*tˆ2)
2. (1 point)
Find the velocity and position vectors of a particle with ac-
celeration a(t) = 9k, and initial conditions v(0) = 3j+1kand
r(0) = 5i−4j−4k.
v(t) = i+j+k
r(t) = i+j+k
Solution:
SOLUTION
a(t) = 9k⇒v(t) = R(9k)dt =9tk+Cand 3j+1k=
v(0) = C,
so C=3j+1kand v(t) = 3j+ (1+9t)k
r(t) = R(3j+ (1+9t)k)dt =3ti+ (1t+ (9/2)t2)k+D.
But 5i−4j−4k=r(0) = D, so D=5i−4j−4kand
r(t) = 5i+ (3t−4)j+ (1t+ (9/2)t2−4)k
Answer(s) submitted:
•0
•3
•9t+1
•5
•-4+3t
•4.5tˆ(2)+1t-4
(correct)
Correct Answers:
•0
•3
•9*t + 1
•0*t + 5
•3*t + -4
•9*t*t/2 + 1*t + -4
3. (1 point) Given that the acceleration vector is a(t) =
(−25cos(5t))i+(−25sin(5t))j+(−2t)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
= (−5sin(5t)))i+ (5cos(5t)) j+−1t2k+C
and i+k=v(0) = 5j+C, so C=i−5j+kand
v(t) = (1−5sin(5t))i+ (5cos(5t)−5)j+−1t2+1k
B.
r(t) = Rv(t)dt
= (cos(5t) +t)i+ (sin(5t)−5t)j+−1
3t3+tk+D.
But i+j+k=r(0) = i+D, so D=j+kand
r(t) = (cos(5t) +t)i+ (sin(5t)−5t+1)j+−1
3t3+t+1k
Answer(s) submitted:
•(-5sin+(5t))+1
•5cos+(5t)-5
•-1tˆ(2)+1
•cos(5t)+t
•sin(5t)-(5t)+1
•(((-1t)ˆ(3))/3)+t+1
(correct)
Correct Answers:
•1-5*sin(5*t)
•5*cos(5*t)-5
•1-2*tˆ2/2
•cos(5*t)+t
•sin(5*t)-5*t+1
•t-2*tˆ3/6+1
1
4. (1 point) The position function of a particle is given by
r(t) = h−2t2,−4t,t2−2ti.
At what time is the speed minimum?
Solution:
SOLUTION
r(t) = h−2t2,−4t,t2−2ti ⇒ v(t) = h−4t,−4,2t−2i
Speed = |v(t)|=p16t2+16 + (2t−2)2=√20t2−8t+20 .
To detemine when the speed is minimum, we first compute the
derivative of the speed:
d
dt |v(t)|=1
2(20t2−8t+20)−1/2(40t−8).
This is zero if and only if the numerator is zero, that is,
40t−8=0 or t=1
5.
Since d
dt |v(t)|<0 for t<1
5and d
dt |v(t)|>0 for t>1
5, the min-
imum speed is attained at t=1
5.
Answer(s) submitted:
•(1/5)
(correct)
Correct Answers:
•0.2
5. (1 point) A dense particle with mass 2 kg follows the path
r(t) = hsin(7t),cos(6t),2t5/2iwith units in meters and seconds.
What force acts on the mass at t=0?
h,,ikg m/s2
Solution:
SOLUTION
r(t) = hsin(7t),cos(6t),2t5/2i
⇒v(t) = r0(t) = h7cos(7t),−6sin(6t),5t
3
2i
⇒a(t) = v0(t) = −49sin(7t),−36cos(6t),15
2t
1
2
⇒a(0) = h0,−36,0i
By Newton’s Second Law, at t=0,
F(0) = ma(0) = h0,2(−36),0i=h0,−72,0iis the required
force.
Answer(s) submitted:
•0
•-72
•0
(correct)
Correct Answers:
•0
•-72
•0
6. (1 point) A projectile is fired from ground level with an
initial speed of 350 m/sec and an angle of elevation of 30 de-
grees. Use that the acceleration due to gravity is 9.8 m/sec2.
(a) The range of the projectile is meters.
(b) The maximum height of the projectile is
meters.
(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)|=350 and, since the angle of elevation is 30o,
v(0) = 350cos(300)i+350sin(30o)j=175√3i+175j.
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 175√3i+175j=v(0) = C, so v(t) = (175√3)i+ (175 −
9.8t)j.
Integrating again gives r(t) = (175√3t)i+ (175t−4.9t2)j+D
where 0=r(0) = D.
Thus the position function of the projectile is
r(t) = (175√3t)i+ (175t−4.9t2)j
(a) Parametric equations for the projectile are
x(t) = 175√3t,y(t) = 175t−4.9t2.
The projectile reaches the ground when y(t) = 0 ( and t>0)
⇒175t−4.9t2=t(175 −4.9t) = 0⇒t=175
4.9.
So the range is x175
4.9=175√3175
4.9≈10825.3.
(b) The maximum height is reached when y(t)has a critical
number (or, equivalently, when the vertical component of the
velocity is 0): 175 −9.8t=0⇒t=175
9.8.
Thus the maximum height is y175
9.8=175175
9.8−4.9175
9.82≈
1562.5
(c) From part (a), impact occurs at t=175
4.9. Thus, the ve-
locity at impact is v175
4.9= (175√3)i+ (175 −9.8(175
4.9)j=
(175√3)i−175j
and the speed is
p3(175)2+ (175)2=350
Answer(s) submitted:
•10825.31755
•1562.5000
•350
(correct)
Correct Answers:
•10825.3175473055
•1562.5
•350
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.
2
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:
•29.71
(correct)
Correct Answers:
•29.698484809835
8. (1 point) A body of mass 6 kg moves in a (counterclock-
wise) circular path of radius 5 meters, making one revolution
every 8 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.
(,)
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 8 seconds, the
angular frequency is 1
4π.
The position function of the body is then
r(t) = 5cos1
4πt,5sin1
4πt.
The acceleration is then
a(t) = r00(t) = −5
16 π2cos1
4πt,−5
16 π2sin1
4πt.
By Newton’s Second Law,
F=ma(t) = −15
8π2cos1
4πt,−15
8π2sin1
4πt.
The magnitude of the force is 15
8π2.
Answer(s) submitted:
•-(18.5055cos(((pi t)/4)))
•-(18.5055sin(((pi t)/4)))
•18.5055
(correct)
Correct Answers:
•- 6 * 3.92699081698724**2 / 5 * cos(3.92699081698724 * t / 5)
•- 6 * 3.92699081698724**2 / 5 * sin(3.92699081698724 * t / 5)
•18.5055082520425
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