Physics 590 Homework, Week 3
Week 3, Homework 1
Prob. 3.1.1
A bicyclist climbs a hill at 10 km/h and then descends at constant speed returning
to her starting point. Her average speed for the trip is 16 km/h. What was her
downhill speed?
Reasoning: If the average speed for the entire trip is greater than the constant
speed for the first half of the trip, then the constant speed for the second half of
the trip must be greater than the average speed.
v1=10km/hr * 1000m/km * hr/3600s= 2.8 m/s
v average= 16km/hr * 1000m/km * hr/3600s= 4.4 m/s
v average= (v1 + v2)/2
4.4 m/s = (2.8 m/s + v2)/2
8.8 m/s = 2.8m/s – v2
Downhill speed = 6.0 m/s
6.0 m/s = v2
This result is consistent with my reasoning.
Prob. 3.1.2
A train accelerates at a constant 0.68 m/s2, starting from an initial speed of 50
km/h. (a) How long does the train take to reach 85 km/h? (b) What distance does
the train cover during the acceleration from 50 km/h to 85 km/h?
a)
To get the time I can use the given variables in the formula: v= v0 + at
v0 = 59 km/hr = 16.4 m/s
v = 85 km/hr = 23.6 m/s
a = 0.68 m/s/s
23.6m/s = 16.4m/s + (0.68m/s/s)t
23.6m/s - 16.4m/s = (0.68m/s/s)t
7.2m/s = (0.68m/s/s)t
(7.2m/s)/(0.68m/s/s) = t
t= 10.588s
a) time for train to reach 85 km/hr = 11s
b)
To calculate the distance I use the formula: x = [(v0 + v)/2]*t
v0 = 59 km/hr = 16.4 m/s
v = 85 km/hr = 23.6 m/s
t = 11s
x= [(16.4m/s + 23.6 m/s)/2]*11s
x= [(40m/s)/2]*11s
x= 20m/s * 11s
x= 220 m
b) distance covered = 220 m
Prob. 3.1.3
You throw a stone vertically upward at 20 m/s from a cliff that is 100 m high. (a)
When does the stone return to the level at which it left your hand (presumed to
be at the height of the cliff)? (b) What distance does the stone travel in reaching
the ground assuming it just misses the edge of the cliff?
a) I assume air resistance is ignored in this problem. I can easily solve for
the time it will take to reach the top of its flight by using the velocity as
zero at the highest point. The time to the highest point will be equal to the
time it will take to return to its original level. As a result; doubling the time
from to the maximum height will give me the total time elapsed for the
stone to return to its original level.
v0 = 20 m/s
v (top of flight) = 0m/s
a = -9.8m/s/s
v= v0 + at
0m/s = 20m/s + (-9.8m/s/s)t
-20 m/s = (-9.8m/s/s)t
(-20m/s)/(-9.8m/s/s) = t
t = 2.0s
Time to return to original level = 2t
2*2.0 = 4.0s
a) Time to return to original level = 4.0s
b) To solve for the total distance the stone travels to the bottom of the cliff, I
need to solve for the distance to the maximum height, double it, and add
that to the height of the cliff. I have already solved for the time to reach
the maximum height, and I can use that to calculate the distance from the
top of the cliff to the top of its flight.
v0 = 20 m/s
v (top of flight) = 0m/s
t = 4.0 s
y = [(v0 + v)/2]*t
y= [(20m/s + 0m/s)/2](2.0s)
y= [(20m/s)/2](2.0s)
y= (10m/s)(2.0s)
y= 20m
Height of cliff + 2 * height of throw = distance
100m + 2(20m) = 140m
b) total distance the stone travels = 140m
Prob. 3.1.4
Describe in words the motion of a remote-controlled car where the graph below
results from the sonic ranger reading of the car’s position vs. time. Be careful to
have your words match, quantititatively, every aspect of the graph, e.g. the
starting and ending positions for various motions, the velocity (magnitude and
direction!) and the acceleration for various parts of the graph (if any).
1) The car starts out moving away from the starting point at about 2m/s and
undergoes constant deceleration to about 0m/s in the time interval from
0.0s to 1.0s.
2) The car travels at constant speed away from the starting point at about
0.33m/s in the time interval from 1.0s to 3.0s.
3) The car stops 1.5m from the starting point for 0.5s, in the time interval
from 3.0s to 3.5s.
4) The car moves away from the starting point under constant acceleration
from about 0m/s to about 2m/s in 0.5s, in the time interval from 3.5s to
4.0s.
5) The car then moves under constant deceleration from about 2m/s to 0m/s
in 0.5s, from the time interval 4.0s to 4.5s.
6) At the 4.5s mark the car changes direction and moves back toward the
starting point traveling under constant acceleration from 0m/s to -2m/s, in
the time interval from 4.5s to 5.0s.
7) The car returns back to the starting point under constant velocity at -2m/s,
in the time interval from 5.0s to 6.0s.
Prob. 3.1.5
A flower pot is dropped, from rest, from the roof of a building. The windows of the
building are of length h and spaced a distance h apart. If the pot takes a time t to
pass from the top to the bottom of the top floor window, find the time it will take to
fall from the top to the bottom of the second window.
Reasoning: I am going to use the formula: y= v0*t + (1/2)at^2 to solve for the
time elapsed to get from the top of the building to the bottom of two different but
adjacent h distances from the top of the building. By subtracting the time of the
higher h section from the lower h section, it will give the time interval that the
flower pot took to cover the distance. I will try various distances for h and
compare the ratios between the time elapsed crossing the first window to the
time elapsed passing the second window. If the ratios are equal, then that ratio
multiplied by t will equal the time it will take to cross the second window.
a) The first value I will use for h is 1.0m
I rearrange the formula:
h= v0*t + (1/2)at^2
h = 0m/s, so the second term is equal to zero.
h = (1/2)at^2 which is rearranged to:
[(2h)/(9.8m/s/s)]^1/2 = t
Time to cross 1h from top of building:
[(2*1m)/(9.8m/s/s)]^1/2 = t
t= .45175
time to cross 2h from the top of the building = .63888s
time to cross window one: .63888s - .45175s = .187s
time to cross 3h from top of building= .782s
time to cross 4h from top of building- .904s
time to cross window two: .904s - .782 = .122s
ratio from window 2 to window 1: .122s/.187s = .65
b) The second value I will use for h is 5.0m
Time to cross 1h from top of building: 1.01s
Time to cross 2h from top of building: 1.43s
time to cross window one: 1.43s – 1.01s = 0.42s
Time to cross 3h from top of building: 1.75s
Time to cross 4h from top of building: 2.02s
time to cross window two: 2.02s – 1.75s = .27s
ratio from window 2 to window 1: 0.27s/0.42s = .64
c) The third value I will use for h is 75m
Time to cross 1h from top of building: 3.91s
Time to cross 2h from top of building: 5.53s
time to cross window one: 5.53s – 3.91s = 1.62s
Time to cross 3h from top of building: 6.78s
Time to cross 4h from top of building: 7.82s
time to cross window two: 7.82s – 6.78s = 1.04s
ratio from window 2 to window 1: 1.04s/1.62s = .64
By trying multiple values for h and always coming to the same ratio between
window 2 and window 1, t multiplied by .64 will give the time to pass window 2
o.64t = the time to pass window 2
Prob. 3.1.6
A commuting train travels between two stations located 1.8 km apart. The train
accelerates for half the distance and decelerates for the remaining half of the
distance. Assume that the magnitude of the acceleration in both cases is 1.2
m/s2. (a) What is the maximum speed achieved by the train? (b) How long does
the trip between stations take?
a)
The maximum speed of the train will be at the halfway point. The train
undergoes constant acceleration until that instant, and will undergo constant
deceleration from that point on until it stops.
v0 = 0m/s
a = 1.2m/s/s
d = 900m
v2 = (v0) + 2*a*(x – x0)
v^2 = (0m/s)^2 + 2(1.2m/s/s)(900m – 0 m)
v^2 = 2.4m/s/s * 900m
v^2 = 2160m^2/s^2
v = 46.4738m/s
a) maximum speed = 46m/s
b) It will take the train the same amount of time to stop from the midpoint of
the trip; given that the acceleration is equal in magnitude and opposite in sign,
and the distances are equal. I will find the time for the first half of the trip and
multiply it by 2 to get the total time for the trip.
v0 = 0m/s
v = 46m/s
a = 1.2m/s/s
v= v0 + at
46m/s = 0m/s + (1.2m/s/s)t
46m/s = (1.2m/s/s)t
(46m/s)/(9.8m/s/s) = t
t = 38.33s
2t=time for entire trip
2t = 76.66s
a) time for entire trip = 77s
Prob. 3.1.7
If Galileo dropped a lead weight from the top of the Leaning Tower of Pisa (which
is 54.6 m high), (a) how fast is the lead weight moving when it hits the ground?
(b) how long does it take the weight to travel the last third of the total 54.6
meters? C) How does this compare to the time needed to travel the first third of
the total distance?
a)
First I will solve for the time it takes to get to the ground, and use that to calculate
the velocity.
y= v0*t + (1/2)at^2
v0 = 0m/s
a = -9.8m/s/s
y = -54.6m
54.6m = 0m/s(t) + ½(-9.8m/s/s)t^2
54.6m + - 9.8m/s/s * t^2
(54.6)/(-9.8m/s/s) = t^2
(11.143)^1/2 = t
t = 3.34s
v= v0 + at
v0 = 0m/s
a = -9.8m/s/s
t = 3.34s
v = 0m/s + (-9.8m/s/s)(3.34s)
v= -32.7m/s
a) velocity at the ground = -32.7m/s
b)
I divide the tower height by 3 to calculate the distance for the final 1/3 of the
tower. 54.6m/3 = 18.2m Each 1/3 section of the tower is 18.2m in length.
I calculate the time it takes to drop 2/3 of the way down (36.4m) and subtract
it from the total time for the drop.
y= v0*t + (1/2)at^2
v0 = 0m/s
a = -9.8m/s/s
y = -36.4m
-36.4m = 0m/s * t + ½(-9.8m/s/s)t^2
(-72.8m)/(-9.8m/s/s) = t^2
((7.35s^2)^1/2 = t
t = 2.72s
t total – t 2/3 = 3.34s – 2.72s = 0.67s
t last 1/3 = 0.67s
b) time to fall the final 1/3 = 0.67s
c) First I need to calculate the time for the first 1/3 down the tower, and then
compare it to the last 2/3 of the tower.
y= v0*t + (1/2)at^2
v0 = 0m/s
a = -9.8m/s/s
y = -18.2m
-18.2m = 0m/s * t + ½(-9.8m/s/s) * t^2
[(-36.4m)/(-9.8m/s/s)^1/2 = t
t = 1.92s
I will compare them by calculating how many times longer it takes to travel the
first 1/3 to the last 1/3.
(1.92s)/(0.67s) = 2.9
c) It takes the object almost 3 (2.9 times longer was my calculated value)
times longer to fall the first 1/3 of the tower than the last 1/3 of the tower
Prob. 3.1.8
An irresponsible physics student at the top of DRL (height of 16.5 m) tries to
bean the professor by dropping their dessert (chocolate pudding) on the
professor (height of 1.90 m) walking on the sidewalk below. The student figures
that accurately calculating the initial speed needed for the pudding will mitigate
the otherwise inevitable negative consequences of a beaning attempt. The trick
is that the professor is obscured by a tree until he emerges just 3.00 meters
horizontally away from the spot directly underneath the student. If the professor
is walking at a constant 1.33 m/s, find the initial velocity needed for the pudding.
Remember that misses will be spotted and lead to those “negative
consequences”. Assume the path the professor is walking is perfectly horizontal.
Reasoning: The first thing that needs to be calculated is the time until the
professor is directly under the student. The next step is to then find out how long
it will take for an object to free fall from the roof to the to a point 1.9m from the
ground. If it takes longer for the pudding to hit that point than the professor to get
there, then it must be thrown down with an initial velocity to compensate for the
difference in time. If the professor takes longer to get to the spot than the
pudding in free fall, then the pudding must be thrown upward to compensate for
the difference in time.
t= d/v
v (prof.) = 1.33m/s
d= 3.00m
t= (3.00m)/(1.33m/s) = 2.2556s
Calculate time for free fall
v0 = 0m/s
y0 = 16.5m
y = 1.9m
a = -9.8m/s/s
y= v0*t + (1/2)at^2
(1.9m - 16.5m) = 0m/s * t + ½(-9.8m/s/s)*t^2
-14.6m = -4.9m/s/s * t^2
(-14.6m)/(-4.9m/s/s) = t^2
(2.97s^2)^1/2 = t
t = 1.7261s = time for free fall
Since the professor takes longer to get to that point, the pudding must be thrown
upward to make up the difference in time.
Time of professor – time of free fall = time the pudding must be thrown up and
return to the original height.
t(p)= 2.2556s
t(pudding)= 1.7261s
2.2556 – 1.7261 = 0.49464s
By dividing the time by 2, I get the time to the top of the toss upward, where v
equals zero. With that data I can calculate the initial velocity.
v = 0m/s
t = (0.49464s)/2 = 0.2473s
a = -9.8m/s/s
v= v0 + at
0m/s = v0 + (-9.8m/s/s)(0.2473)
v0 = 2.424 m/s
V initial = 2.42 m/s (upward)
Prob. 3.1.9
A stone is dropped into the water from a bridge 144 ft above the water. Another
stone is thrown vertically down 1.0 s after the first is dropped. Both stones strike
the water at the same time. (a) What was the initial speed of the second stone?
(b) Plot the speed versus time on a graph for each stone taking zero time as the
instant the first stone was released.
Reasoning: First I need to solve for the time for the first stone to hit the water.
Then subtract 1 from that time to get the time the second stone needs to cross
the same distance. From there I can solve for the second stone’s initial velocity.
y0 = 144ft = 43.9m
y = 0m
v0 = 0m/s
a = -9.8m/s/s
y= v0*t + (1/2)at^2
(0m – 43.9m) = 0m/s * t = ½(-9.8m/s/s) * t^2
-43.9m = (-4.9m/s/s) * t^2
[(-43.9m)/(-4.9m/s/s)]^1/2 = t
t = 3.0s
Time for second stone = time of first stone – 1.0s
t(2nd) = 2.0s
y0 = 144ft = 43.9m
y = 0m
t = 2.0s
a = -9.8m/s/s
y= v0*t + (1/2)at^2
(0m – 43.9m) = v0 * (2.0s) + ½(-9.8m/s/s)(2.0s)^2
-43.9m = v0 * (2.0s) – 19.6m
19.6m – 43.9m = v0 * (2.0s)
-24.3m = v0 * (2.0s)
(-24.3m)/(2.0s) = v0
v0 = -12.15m/s
V initial for the second rock is -12.2m/s (down)
Graph of stone 1
Velocity vs Time (acceleration) stone # 1
0
Velocity (m/s)
-5 0
1
2
3
4
-10
-15
Series1
-20
-25
-30
-35
Time (s)
Graph of stone 2
Velocity vs Time (acceleration) stone # 2
0
Velocity (m/s)
-10 0
1
2
3
4
-20
-30
Series1
-40
-50
-60
-70
Time (s)
Graph of stone and two together
Velocity vs time (stone 1 & 2
0
Velocity (m/s)
-10 0
0.5
1
1.5
2
2.5
3
3.5
-20
-30
-40
-50
-60
-70
Time (s)
Both stones are falling with the same acceleration, even though they started with
different initial velocities. The slopes are identical for both stones, and that
shows that their velocities were changing by the same amount over the same
period of time.
Prob. 3.1.10
An elevator ascends with an upward acceleration of 1.5 m/s2. At the instant the
upward speed is 2.5 m/s, a loose bolt drops from the ceiling of the elevator 3
meters from the floor. Calculate (a) the time of flight of the bolt from the ceiling to
the floor and (b) the distance it has fallen relative to the elevator shaft.
Reasoning: The bolt will fall down and the elevator rises. The point at which they strike
each other relative to the elevator shaft will coincide with the time it will take for the
floor and bolt to make contact. By plugging in times for t, I will find the distance the
coin will fall and the distance the elevator will rise. When the distances combined equal
-3.0 m, (by the equation y(coin) – y(elevator) that will equal the time of impact. If the
combined distances are greater (less negative or positive) than -3.00m, then the time of
impact has been passed. If the combined times are less than -3.00m (more negative), then
the impact has not yet occurred. This will also give us a distance relative to where the
elevator floor started out in relation to the elevator shaft, because the distance the floor
rose is solved for in the equation: y= v0*t + (1/2)at^2
a)
Coin (1 s):
Xc0= 0m
Vc0 = 2.5m/s
A = -9.8m/s/s
t = 1.00s
y= v0*t + (1/2)at^2
y = (2.5m/s)(1.00s) + ½(-9.8m/s/s)(1.00s)^2
y = - 2.40m
Elevator (1 s):
Xe0= 0m
Ve0 = 2.5m/s
A = 1.5m/s/s
t = 1.00s
y= v0*t + (1/2)at^2
y = (2.5m/s)(1.00s) + ½(1.5m/s/s)(1.00s)^2
y = 3.25m
y(coin) – y(elevator) will equal the time of impact
-2.40m – (-3.25m) = .85m
I now know 1 second is too long of a time. I will try the time at .25 seconds.
Coin (1 s):
Xc0= 0m
Vc0 = 2.5m/s
A = -9.8m/s/s
t = 0.25s
y= v0*t + (1/2)at^2
y = (2.5m/s)(0.25s) + ½(-9.8m/s/s)(0.25s)^2
y = 0.31875m
Elevator (0.25 s):
Xe0= 0m
Ve0 = 2.5m/s
A = 1.5m/s/s
t = 0.25s
y= v0*t + (1/2)at^2
y = (2.5m/s)(0.25s) + ½(1.5m/s/s)(0.25s)^2
y = 0.8125m
0.31875m – (.8125) = -.494m
That time is too short, I will try 0.50s
Coin (0.50s):
Xc0= 0m
Vc0 = 2.5m/s
A = -9.8m/s/s
t = 0.50s
y= v0*t + (1/2)at^2
y = (2.5m/s)(0.50s) + ½(-9.8m/s/s)(0.50s)^2
y = 0.03m
Elevator (0.50s):
Xe0= 0m
Ve0 = 2.5m/s
A = 1.5m/s/s
t = 0.50s
y= v0*t + (1/2)at^2
y = (2.5m/s)(0.50s) + ½(1.5m/s/s)(0.50s)^2
y = 1.40m
0.03m – (1.40) = -1.37m
This time is also too short, but getting closer. I will try 0.75s
Coin (0.75s):
Xc0= 0m
Vc0 = 2.5m/s
A = -9.8m/s/s
t = 0.75s
y= v0*t + (1/2)at^2
y = (2.5m/s)(0.75s) + ½(-9.8m/s/s)(0.75s)^2
y = -0.88mm
Elevator (0.75 s):
Xe0= 0m
Ve0 = 2.5m/s
A = 1.5m/s/s
t = 0.75s
y= v0*t + (1/2)at^2
y = (2.5m/s)(0.75s) + ½(1.5m/s/s)(0.75s)^2
y = 2.297m
-0.88m – (2.297) = -3.18m
Now I have just a little too much time, I will try 0.73s
Coin (0.73s):
Xc0= 0m
Vc0 = 2.5m/s
A = -9.8m/s/s
t = 0.73s
y= v0*t + (1/2)at^2
y = (2.5m/s)(0.73s) + ½(-9.8m/s/s)(0.73s)^2
y = -0.78621m
Elevator (0.73 s):
Xe0= 0m
Ve0 = 2.5m/s
A = 1.5m/s/s
t = 0.73s
y= v0*t + (1/2)at^2
y = (2.5m/s)(0.73s) + ½(1.5m/s/s)(0.73s)^2
y = 2.2246m
-0.78621m – (2.2246) = -3.01m
The time for the bolt to hit the floor is 0.73s
The distance at which this happened relative to a point on the elevator shaft that was
parallel to the floor at time zero, is 2.22 m above that point.