PSCI 1055
Homework #3 KEY
Spring 2008
Buckley
This homework is due on Wednesday, January 30, 2008, at classtime. Late homework is
not accepted.
1. (12 points) Consider the trip depicted below. Each grid corresponds to one mile on a
side. The time of the trip on each segment is given in the table below. Assume
compass directions are normal – up is north, down is south, left is west, and right is
east. Fill in the remainder of the table.
F
E
B
A
C
D
Segment
Distance
(miles)
2.0
Displacement
(miles)
2.0 N
Average speed
Average velocity
A
Time
(min)
4.0
2.0/4.0 = 0.50 mpm
2.0 N/4.0 = 0.50 mpm N
B
5.0
3.0
3.0 E
3.0/5.0 = 0.60 mpm
3.0 E/5 = 0.60 mpm E
C
2.5
3.0
3.0 S
3/2.5 = 1.2 mpm
3.0 S/2.5 = 1.2 mpm S
D
6.0
2.0
2.0 W
2.0/6.0=0.33 mpm
2.0 W/6.0=0.33 mpm W
E
15.0
7.0
7.0 N
7.0/15.0=0.47 mpm
7.0/15.0=0.47 mpm N
F
7.5
5.0
5.0 E
5.0/7.5 = 0.67 mpm
5.0 E/7.5=0.67 mpm E
³OVERº
2. (10 points)
a. A ball is dropped from the top of a building with an initial velocity of zero. If the
ball takes 3.5 s to hit the ground, how tall is the building?
This problem asks for a distance and gives a time and an initial velocity. We
also know that since it is a free fall problem the acceleration due to gravity is
9.8 m/s2. (Notice if we did this from a planet on the moon the answer would
come out differently since the acceleration due to gravity would be different.)
Sorting through our equations, the one that relates distance to time and
acceleration is probably most appropriate. (d = ½ at2)
d = 1 at 2 = 1 (9.8 m / s 2 )(3.5s ) 2 = 60 m
2
2
b. How fast will the ball of part a be traveling when it hits the ground?
Here we are looking for a velocity (speed) when the ball hits the ground and
are given the initial velocity (0 m/s from part a), the time it falls (3.5 s from
a), the acceleration due to gravity (9.8 m/s2) and the distance it falls (60 m
from a). The equation that will be able to give us the final velocity is: vf = vo
+ at. Using the information from the problem:
v f = vo + at = 0 m / s + (9.8m / s 2 )(3.5 s ) = 34 m / s
c. A ball is thrown into the air with an initial vertical velocity of 5.0 m/s. How long
will it be before the ball reaches its peak and starts dropping back to the Earth?
This problem is a projectile problem. In projectile problems we typically
separate the vertical component from the horizontal component. The
vertical component can tell us things like the length of time something is in
the air, how high it goes, etc. The vertical component is subject to the
acceleration due to gravity. The horizontal component can tell how far
horizontally something travels while in the air – it is important to realize
there is no acceleration in the horizontal direction. Once the object has a
horizontal velocity, we assume that component remains the same for the
whole trip.
In this question, we are looking for the time the ball takes to reach its peak –
this means we are interested in the horizontal component. We have the
initial velocity (5.0 m/s), the final velocity (0 m/s at the peak), and the
acceleration due to gravity. The equation vf = vo + at is our best candidate.
Remember the acceleration due to gravity and the velocity are in opposite
directions since the ball is traveling up and gravity is accelerating it downward.
This means they will have opposite algebraic signs.
v f = vo + at ⇒ 0m / s = 5.0m / s + (−9.8m / s 2 )t ⇒ −5.0m / s = (−9.8m / s 2 )t ⇒ t =
−5.0m / s
= 0.51s
−9.8m / s 2
d. If the ball in part c is thrown with a horizontal velocity component of 25.0 m/s in
addition to its vertical velocity of 5.0 m/s, how far would the ball travel
horizontally before it reached its peak?
Remember there is no acceleration in the horizontal direction – the ball
travels 25.0 m/s horizontally until it hits the ground again. Thus, from part
c, the ball is in the air for 0.51 s. A ball traveling at 25.0 m/s for 0.51 s will
travel 25.0 m/s x 0.51 s = 12.8 m to reach the peak.
e. How far would the ball of part d travel before it hit the ground?
If it takes 0.51 s to reach the peak, it will take another 0.51 s to come back to
where it started. Total travel time is 1.02 s x 25.0 m/s = 25.5 m
3. (2 points) The centripetal acceleration is a vector quantity. Which way does the
centripetal vector point for an object traveling in a circle?
Centripetal means “center-seeking” and that is the direction the centripetal
acceleration (and force) vector will point.
4. (2 points) If the speed of an object is doubled, what effect does that have on the
centripetal acceleration – halved, doubled, tripled, quadrupled, etc.? Explain your
reasoning.
v2
so a doubling of the speed will lead to
r
an increase in centripetal acceleration of a factor of (22) = 4.
Centripetal acceleration is given by ac =