PUM Physics II - Kinematics
Lesson 15 Solutions (corrected)
15.1 Derive (very challenging!)
v(t) = vo + at
x(t) = xo + vot + ½at2
Solve for t using first equation:
v = vo + at
t = (v – vo)/a
Substitute expression for t into second equation:
x = xo + vot + ½at2
x = xo + vo[(v – v0)/a] + ½a[(v – vo)/a]2
x = xo + (vov – vo2)/a + (v2 – 2vov + vo2)/2a
x – xo = (2vov – 2vo2)/2a + (v2 – 2vov + vo2)/2a
x – xo = (2vov – 2vo2 + v2 – 2vov + vo2)/2a
x – xo = (v2 – vo2)/2a
v2 = vo2 + 2a(x-xo)
15.2 Practice
a) vo = 10 m/s
v = 0 m/s
a = -10 m/s2
xo = 0 m
v2 = vo2 + 2a(x–xo)
(v2 – vo2)/2a = x-xo
x = (-10 m/s)2/2(-10 m/s2) = 5.0 m
b) vo = 0 m/s
xo = 1.5 m
a = -10 m/s2
x=0m
v2 = vo2 + 2a(x–xo)
v = [vo2 + 2a(x–xo)]1/2
-v = [2(-10 m/s2)(-1.5 m)]1/2 = -5.5 m/s
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PUM Physics II - Kinematics
Lesson 15 Solutions (corrected)
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c) vo = 15 m/s
v = 10 m/s
x-xo = 200 m
v2 = vo2 + 2a(x–xo)
a = (v2 – vo2) /[2(x–xo)]
a = [(10 m/s)2 – (15 m/s)2]/[2(200 m)] = -0.31 m/s2
d) vo = 200 m/s
v = 0 m/s
x-xo = 40 cm = 0.4 m
v2 = vo2 + 2a(x–xo)
a = (v2 – vo2) /[2(x–xo)]
a = – (200 m/s)2/[2(0.4 m)] = -5x10 m/s2
15.3 Represent and reason
a) Middle set of graphs.
b) Right set of graphs. Middle one initially slows before speeding up.
c) Right set of graphs.
Position (m)
Time (sec)
Velocity (m/s)
Time (sec)
(m/s2)
Accel.
Accel.
(m/s2)
Time (sec)
Time (sec)
Time (sec)
Time (sec)
Accel.
Velocity (m/s)
Velocity (m/s)
Time (sec)
Time (sec)
(m/s2)
Position (m)
Position (m)
15.4 Represent and Reason
Time (sec)
PUM Physics II - Kinematics
Lesson 15 Solutions (corrected)
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Left Graph: Object is moving in the negative direction at constant negative velocity then reverses
direction abruptly and starts moving in the positive direction at constant positive velocity.
Middle Graph: Object is initially moving very fast in the positive direction, but quickly slows
down before traveling at a constant positive velocity.
Right Graph: Object starts at a negative position traveling slowly in the positive direction. Object
speeds up until reaching a constant positive velocity.
PUM Physics II - Kinematics
Lesson 15 Solutions (corrected)
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15.5 Summarize
Describe the Motion…
In words and provide an
example.
Motion at Constant Velocity
The object’s velocity stays constant
meaning it covers the same distance
every second.
An example would be a ball rolled
across a smooth level track.
Motion at Constant Acceleration.
The object's velocity is increasing or
decreasing by the same amount every
second.
For example, a cart going down a smooth
track that is tilted at an angle.
Speeding up:
With a motion diagram.
v1
v2
Δv12
v3
v1
Δv23
Δv12
x
v2
v3
Δv23
x
With a graph of position
versus clock reading.
t
Mathematically as a function
x(t).
t
x(t) = xo + vo(t) + ½a(t)2
x(t) = xo + vo(t)
PUM Physics II - Kinematics
Describe the Motion…
Lesson 15 Solutions (corrected)
Motion with Constant Velocity
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Motion with Constant Acceleration.
v
v
With a graph of velocity
versus clock reading.
0
t
t
Mathematically as a function
v(t).
v(t) = constant
v(t) = vo + at
a
a
With a graph of acceleration
versus clock reading.
t
t
Mathematically as a function
a(t).
a(t) = 0
Homework
15.6 Regular Problem
vo = -13.7 m/s
v = 0 m/s
a = 4.1 m/s2
v2 = vo2 + 2a(x–xo)
Δx = (v2 – vo2)/(2a)
 –  13.7 m / s 2 
   22.9 m
x  
2
 2  4.1 m / s  


a(t) = constant
PUM Physics II - Kinematics
Lesson 15 Solutions (corrected)
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15.7Represent and Reason
II
III
Position (m)
Time (sec)
Velocity (m/s)
Time (sec)
Accel.
(m/s2)
Time (sec)
Accel.
(m/s2)
Time (sec)
Time (sec)
Time (sec)
Accel.
Velocity (m/s)
Velocity (m/s)
Time (sec)
Time (sec)
(m/s2)
Position (m)
Position (m)
I
Time (sec)
a) III
b) I
c) II
15.8 Evaluate the Solution
We can’t assume that the acceleration is 9.8m/s2 since she is holding onto the pole. We can
determine the acceleration first:
x(t) = xo + vot + ½at2
0 m = 2.0 m + ½a(2.0 s)2
a = [2(-2.0 m)]/(2.0 s)2 = -1.0 m/s2
Now we can determine her final velocity:
v = v0 + at = (-1.0 m/s2)(2.0 s) = -2.0 m/s
PUM Physics II - Kinematics
Lesson 15 Solutions (corrected)
a = (v2 – vo2) /[2(x–xo)] = [(0 m/s)2 – (-2.0 m/s)2]/[2(0.10 m)] = 20 m/s2
This is a far more reasonable answer then 1920.8 m/s2
15.9Regular Problem
Find the initial velocity:
v2 = vo2 + 2a(x–xo)
(0 m/s)2 = vo2 + 2(-10 m/s2)(13 m – 2.1 m)
- vo2 = 2(-10 m/s2)(10.9 m)
-vo = [2(-10 m/s2)(10.9 m)]1/2 = 14.8 m/s
2.5 s later:
x(t) = (2.1 m) + (14.8 m/s)t + ½(-10 m/s2)t2
x = (2.1 m) + (14.8 m/s)(2.5 s) + ½(-10 m/s2)(2.5 s)2 = 7.9 m
v(t) = (14.8 m/s) + (-10 m/s2)t
v = (14.8 m/s) + (-10 m/s2)(2.5 s) = -10.2 m/s
15.10 Regular Problem
Assuming that Komila and Heather are moving at constant velocity:
7 mi/hr = 3.1 m/s
5 mi/hr = 2.2 m/s
x(t)Heather = (3.1 m/s)t
x(t)Komila = (50 m) + (2.2 m/s)t
When will they meet?
x(t)Heather = x(t)Komila
(3.1 m/s)t = (50 m) + (2.2 m/s)t
t = (50 m)/(0.9 m/s) = 55.6 s
15.11 Regular Problem
Assuming that Komila and Heather are moving at constant velocity:
x(t)Heather = (3.1 m/s)t
x(t)Komila = (50 m) + -(2.2 m/s)t
When will they meet?
x(t)Heather = x(t)Komila
(3.1 m/s)t = (50 m) + (-2.2 m/s)t
t = (50 m)/(0.9 m/s) = 55.6 s
t = (50 m)/(5.3 m/s) = 9.4 s
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PUM Physics II - Kinematics
Lesson 15 Solutions (corrected)
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15.12 Reason
Lets say two cars collide with each other, one going at 50 mi/hr to the right and the other going
40 mi/hr to the left. Using the idea of relative motion this is equivalent to one car hitting a parked
car at 90 mi/hr. The change in velocity will be much greater and thus the acceleration will be
much greater, making the crash more dangerous.
On the other hand lets say two cars collide with each other one going 50 mi/hr to the right while
the other car is going 40 mi/hr to the right as well. Again using the idea of relative motion, this is
like one car hitting a parked car at 10 mi/hr. The change in velocity is much smaller and thus the
acceleration is much smaller, making the crash less dangerous.