Larena Whitney-krug Platte MAT 267 ONLINE A Summer 2021
Assignment Section 10.9 due 05/26/2021 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
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) = 7k, and initial conditions v(0) = 5j−2kand
r(0) = 5i+3j+2k.
v(t) = i+j+k
r(t) = i+j+k
Solution:
SOLUTION
a(t) = 7k⇒v(t) = R(7k)dt =7tk+Cand 5j−2k=
v(0) = C,
so C=5j−2kand v(t) = 5j+ (−2+7t)k
r(t) = R(5j+ (−2+7t)k)dt =5ti+ (−2t+ (7/2)t2)k+D.
But 5i+3j+2k=r(0) = D, so D=5i+3j+2kand
r(t) = 5i+ (5t+3)j+ (−2t+ (7/2)t2+2)k
Correct Answers:
•0
•5
•7*t + -2
•0*t + 5
•5*t + 3
•7*t*t/2 + -2*t + 2
3. (1 point) Given that the acceleration vector is a(t) =
(−(cos(t)))i+ (−(sin(t)))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
= (−(sin(t))))i+ (cos(t))j+1t2k+C
and i+k=v(0) = 1j+C, so C=i−1j+kand
v(t) = (1−sin(t))i+ (cos(t)−1)j+1t2+1k
B.
r(t) = Rv(t)dt
= (cos(t) +t)i+ (sin(t)−t)j+1
3t3+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
3t3+t+1k
Correct Answers:
•1-sin(t)
•cos(t)-1
•2*tˆ2/2+1
•cos(t)+t
•sin(t)-t+1
•2*tˆ3/6+t+1
4. (1 point) The position function of a particle is given by
r(t) = h2t2,1t,t2−3ti.
At what time is the speed minimum?
Solution:
SOLUTION
r(t) = h2t2,1t,t2−3ti ⇒ v(t) = h4t,1,2t−3i
Speed = |v(t)|=p16t2+1+ (2t−3)2=√20t2−12t+10 .
To detemine when the speed is minimum, we first compute the
derivative of the speed:
d
dt |v(t)|=1
2(20t2−12t+10)−1/2(40t−12).
This is zero if and only if the numerator is zero, that is,
40t−12 =0 or t=3
10 .
Since d
dt |v(t)|<0 for t<3
10 and d
dt |v(t)|>0 for t>3
10 , the
minimum speed is attained at t=3
10 .
Correct Answers:
•0.3
5. (1 point) A dense particle with mass 7 kg follows the
path r(t) = hsin(9t),cos(6t),2t11/2iwith units in meters and
seconds.
What force acts on the mass at t=0?
h, , ikg m/s2
Solution:
SOLUTION
1
r(t) = hsin(9t),cos(6t),2t11/2i
⇒v(t) = r0(t) = h9cos(9t),−6sin(6t),11t
9
2i
⇒a(t) = v0(t) = −81sin(9t),−36cos(6t),99
2t
7
2
⇒a(0) = h0,−36,0i
By Newton’s Second Law, at t=0,
F(0) = ma(0) = h0,7(−36),0i=h0,−252,0iis the required
force.
Correct Answers:
•0
•-252
•0
6. (1 point) A projectile is fired from ground level with an
initial speed of 600 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)|=600 and, since the angle of elevation is 30o,
v(0) = 600cos(300)i+600sin(30o)j=300√3i+300j.
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 300√3i+300j=v(0) = C, so v(t) = (300√3)i+ (300 −
9.8t)j.
Integrating again gives r(t) = (300√3t)i+ (300t−4.9t2)j+D
where 0=r(0) = D.
Thus the position function of the projectile is
r(t) = (300√3t)i+ (300t−4.9t2)j
(a) Parametric equations for the projectile are
x(t) = 300√3t,y(t) = 300t−4.9t2.
The projectile reaches the ground when y(t) = 0 ( and t>0)
⇒300t−4.9t2=t(300 −4.9t) = 0⇒t=300
4.9.
So the range is x300
4.9=300√3300
4.9≈31813.2.
(b) The maximum height is reached when y(t)has a critical
number (or, equivalently, when the vertical component of the
velocity is 0): 300 −9.8t=0⇒t=300
9.8.
Thus the maximum height is y300
9.8=300300
9.8−4.9300
9.82≈
4591.84
(c) From part (a), impact occurs at t=300
4.9. Thus, the ve-
locity at impact is v300
4.9= (300√3)i+ (300 −9.8(300
4.9)j=
(300√3)i−300j
and the speed is
p3(300)2+ (300)2=600
Correct Answers:
•31813.1780982039
•4591.83673469388
•600
7. (1 point) A ball is thrown at an angle of 45 degrees to the
ground, and lands 10 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 10 meters away, 10 =x v0√2
9.8!=
v0
√2
2
v0√2
9.8=v2
0
9.8, and the initial velocity is
v0=p9.8(10) = √98.
Correct Answers:
•9.89949493661167
8. (1 point) A body of mass 7 kg moves in a (counterclock-
wise) circular path of radius 3 meters, making one revolution
every 10 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.
2
Note: Use exact forms or at least 4 significant digits in your
answers.
Solution:
SOLUTION:
Since the body makes one revolution every 10 seconds, the
angular frequency is 1
5π.
The position function of the body is then
r(t) = 3cos1
5πt,3sin1
5πt.
The acceleration is then
a(t) = r00(t) = −3
25 π2cos1
5πt,−3
25 π2sin1
5πt.
By Newton’s Second Law,
F=ma(t) = −21
25 π2cos1
5πt,−21
25 π2sin1
5πt.
The magnitude of the force is 21
25 π2.
Correct Answers:
•- 7 * 1.88495559215388**2 / 3 * cos(1.88495559215388 * t / 3)
•- 7 * 1.88495559215388**2 / 3 * sin(1.88495559215388 * t / 3)
•8.29046769691506
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