C Kinematics Questions
PART I
26. An object moves along the x-axis such that x = 2 t4 − 3 t2 + 5, where x = position in m and t =
time in s. (a) Determine the velocity function for this object. (b) Determine the acceleration
function for this object. (c) What position reached by the object is closest to the origin? (d)
What is the maximum speed in a leftward direction? (e) What is the greatest magnitude of
acceleration that occurs in the first second of motion?
27. A top fuel dragster can be modeled by d = 0.333 t4 − 7.5 t3 + 91 t2, where d = distance in ft
and t = time in s. The race is ¼ mile or 1320 ft. At the point where it crosses the finish line
determine (a) the time, (b) the speed, and (c) the rate of acceleration. Note: use the features of
your calculator to solve for roots of polynomials – you do not have to solve explicitly.
28. Consider a bungee jumper. The motion that results from the pull of the bungee can be
modeled by the following function: v = −5 t3 + 22 t2 − 9.8 t − 20, where v = velocity in m/s
and t = time in s. This equation is valid only for a certain interval of time starting at the
instant the bungee first starts to pull the falling person. (a) Determine the acceleration as a
function of time. (b) Determine the time at which the bungee stops pulling (hint: the person is
once again in freefall at that point). (c) Find the average acceleration that occurs while the
bungee is pulling. (d) Find the maximum upward acceleration caused by the bungee.
29. (a) Assuming the initial position is y = 0, determine the function describing the position of the
bungee jumper in the previous problem. (b) Find the maximum amount that the bungee cord
is stretched.
30. An object moves along the y-axis in such a way that its acceleration in an upward direction is
given by: a = 5 t − 2, where a is acceleration in m/s and t is time in seconds. At t = 4.0 s, the
velocity of the object is 38 m/s upward and its position is 40.0 above the origin. (a) Find the
object’s velocity at t = 0. (b) Find the object’s position at t = 0. (c) Determine the minimum
speed of the object and the position at which this occurs.
31. Scientists speculate that bacteria may have survived being launched into space on bits of rock
and soil released by the impact of an asteroid hitting the earth’s crust. In such a scenario
bacterium would be subject to a violent motion undergoing a change in acceleration from
zero to 2.00 million g’s in 0.500 ms. Assume the change in acceleration occurs at a constant
rate and that the bacterium is initially at rest. (a) Determine a(t). (b) Determine v(t). (c)
Determine x(t). (d) Find the distance traveled and speed at t = 0.500 ms.
32. Suppose the motion of a rocket can be modeled by the equation a = A + B t0.5. Apply this
model to a rocket launched from rest with initial acceleration 5.5 m/s2 upward that has an
acceleration of 13 m/s2 upward 9.0 seconds after launch. (a) Determine the values of A and
B, including units. (b) Find the distance traveled during the first 9.0 seconds. (c) Find the
speed at 9.0 seconds.
33. An object moves in the xy-plane such that its position in meters is described by x(t) = 4t2 and
y(t) = 3t4 − 10t2, where t is time in seconds. (a) Determine the velocity of the object at t = 2 s.
(b) Determine the acceleration of the object at t = 2 s. (c) At what point in time is the object
moving in a purely horizontal direction. (d) Find the equation of the path traveled by the
object (i.e. find y as a function of x).
34. An object initially at the origin has velocity with components described by vx(t) = 3 − 2t and
vy(t) = 2 + 3t, where v is in m/s and t is in s. (a) Determine the parametric equations that
describe the position of the object as a function of time. (b) Find the acceleration at t = 3 s.
(c) Determine the minimum speed of the object and the position at which this occurs. (d)
Describe the type of path followed and sketch the graph.
35. A wombat is in the back of a flatbed truck unbeknownst to the driver. At the instant the truck
starts moving northward, the wombat runs eastward across the flat bed of the truck and off
the edge. The speed of the truck can be modeled by v = 8 t0.5. Meanwhile the wombat’s
motion is a constant acceleration of 3.0 m/s2 east (relative to the truck) as it crosses the 2.0 m
width of the truck bed. (a) Determine the velocity of the wombat as it flies off the edge of the
truck bed. (b) Determine the acceleration of the wombat at the same instant. (c) Determine
the equation describing the path followed by the wombat. (d) Use the equation of path to
determine how far forward the truck has moved by the time the wombat flies off.
PART II
19. An object moves along the x-axis according to the function x(t) = 5 – 2t + 3t2 – 0.5t3, where t
is time measured in seconds and x is position measured in meters. (a) Determine expressions
for this object’s velocity and acceleration as functions of time. (b) For the interval of time t =
0 to t = 4.0 s determine: (b) average acceleration, (c) maximum rate of acceleration, (d)
average velocity, (e) maximum speed, and (f) maximum distance from the origin.
20. For each of the following expressions determine the indefinite integral with respect to time:
(a) v(t) = 6t3 − 5t
(b) a(t) = 3t2 − 4t + 7
(c) a(t) = 10 + t−2
(d) v(t) = 2t5 − t4/3
(e) v(t) = 12t0.5 + 15
21. Beginning at a position 1.0 m to the right of the origin, an object moves along the x-axis with
velocity in m/s given by v(t) = 0.20t4 – t2, where t is in seconds. (a) Determine its position,
velocity, and acceleration at t = 2.0 s. (b) For t > 0, determine the greatest leftward position,
speed, and acceleration.
22. A certain car is reported by a magazine to complete a “standing quarter mile” in 14.5 seconds
with an ending speed of 95 mph or 139 ft/s. (a) Determine the average acceleration. (b)
Assuming the acceleration is constant calculate the distance the car should travel in 14.5
seconds. How does this compare with one quarter mile (1320 ft)? Explain the discrepancy!
23. The magazine reported data for time and speed of the same car as shown in the table below.
Assuming that speed is proportional to the square root of the time, find an equation for v(t) to
model this data. (a) Use this model to determine the functions x(t) and a(t). (b) Calculate
x(14.5 s) and compare to one quarter mile. (c) The magazine reported the top speed of this
car as 151 mph or 221 ft/s. Determine the acceleration of the car as it reaches 151 mph
according to your functions. What does this result tell us about the functions?
Time (s)
Speed (mph)
Converted Speed (ft/s)
0
0
0
1.9
30
44.0
3.0
40
58.7
4.3
50
73.3
5.9
60
88.0
7.7
70
102.7
10.2
80
117.3
13.0
90
132.0
16.2
100
146.7
20.2
110
161.3
26.9
120
176.0
36.2
130
190.7
24. Try a different approach to the car in the previous example: Assume that the acceleration
decreases linearly such that a(t) = mt + b. (a) Determine values for m and b based on the
known quarter mile (1320 ft) time of 14.5 s and final speed of 139 ft/s. (b) What is the initial
acceleration? (c) Use this new model to determine the top speed of the car.
PART I ANSWERS
26. a. v = 8 t3 − 6 t
b. a = 24 t2 − 6
c. 3.88 m right of origin
d. 2.00 m/s
e. 18 m/s2
27. a. 4.55 s
b. 488 ft/s (333 mph)
c. 60.0 ft/s2
28. a. a = −15 t2 + 44 t − 9.8
b. t = 2.93 s
c. 11.7 m/s2 upward
d. 22.5 m/s2 upward
29. a. y = −1.25t4 + 7.33t3 − 4.9t2 − 20t
b. 22.7 m
30. a. 6.0 m/s, up
b. 21.3 m below origin
c. 5.6 m/s, 19.0 m below origin
31. a. a = 3.92 × 1010 t
b. v = 1.96 × 1010 t 2
c. x = 6.53 × 109 t 3
d. 0.817 m, 4900 m/s
32. a. A = 5.5 ms−2, B = 2.5 ms−2.5
b. 385 m
c. 94.5 m/s
33. a. 58.2 m/s, 74.1°
b. 124 m/s2, 86.3°
c. 1.29 s
d. y = 0.188x2 − 2.5x
34. a. x = 3t − t2, y = 2t + 1.5t2
b. 3.61 m/s2, 123.7°
c. 3.61 m/s at origin
d. parabola w/ vertex on origin but tilted left by 33.7°
35. a. 9.27 m/s, 68.1°
b. 4.78 m/s2, 51.1°
c. y = 3.935x0.75
d. 6.62 m
PART II ANSWERS
19. a. v(t) = −2 + 6t − 1.5 t2
a(t) = 6 − 3t
b. aavg = 0
c. amax = 6 m/s2
d. vavg = 2.00 m/s, right
e. vmax = 4.00 m/s
f. xmax = 13.4 m
20. a. x(t) = 1.5t4 − 2.5t2 + C
b. v(t) = t3 − 2t2 + 7t + C
c. v(t) = 10t − t−1 + C
d. x(t) = t6/3 − t5/15 + C
e. x(t) = 8t1.5 − 15t + C
21. a. x = 0.387 m, left of origin; v = 0.800 m/s, left; a = 2.40 m/s2, right
b. x = 0.4907 m, left; v = 1.25 m/s; a = 1.217 m/s2, left
22. a. 9.61 ft/s2
b. 1010 ft
23. v = 35.06 t 0.5
a. a = 17.53 t −0.5 ; d = 23.37 t 1.5
b. 1290 ft (1/4 mile = 1320 ft)
c. a = 2.78 ft/s2 (should be zero!)
24. a. m = −1.229 ft/s3; b = 18.497 ft/s2
b. 18.5 ft/s2
c. 139.2 ft/s