QUIZ 1
Physics 101, Fall 2008
September 19, 2008
SOLUTIONS
Problem 1. In the future cars may be equipped with devices that can measure the speed of, and
distance to, other cars on the road. Then, the car’s computer might perform calculations such as
the following.
At a moment when car B is 44 m in front of car A, the speed of car A is 88 km/hr and the speed
of car B is 73 km/hr.
a. [2 pts] What are the speeds vA and vB of cars A and B in m/s?
vA = 88
km 1000 m 1 hr
89 m
m
·
·
=
= 24.4 .
hr
1 km 3600 s
3.6 s
s
vB = 73
km 1000 m 1 hr
873 m
m
·
·
=
= 20.3 .
hr
1 km 3600 s
3.6 s
s
b. [3 pts] Car B proceeds with constant speed vB , but car A decelerates for time T at rate a m/s2
so that it just barely avoids colliding with car B. Deduce the time T and the deceleration a > 0.
Take time t = 0 to be the moment that car B is 47 m in front of car A, and when car A begins to
decelerate. Then, during the time while car A decelerates, its speed and position are given by
vA (t) = vA,0 − at,
sA (t) = vA,0 t −
at2
,
2
where VA,0 = 24.4 m/s. During this time, the postion of car B is given by
sB (t) = D0 + vB t,
where D0 = 44 m.
The key to the problem is that at time T (when car A stops decelerating) the speed of car A must be
the same as that of car B, and the position of car A is essentially the same as that of car B. Then, after
this time, the two cars move together.
Therefore
Also,
sA (T ) = sB (T ),
vB = vA,0 − aT,
so that
vA,0 T −
or
D0 =
so that
a=
vA,0 − vB
.
T
aT 2
vA,0 − vB
= vA,0 T −
T = D0 + vB T,
2
2
vA,0 − vB
T.
2
Thus, T = 2D0 /(vA,0 − vB ) = 2 · 44/(24.4 − 20.3) = 21.5 s and a = (vA,0 − vB )/T = 4.1/21.5 = 0.19 m/s2 .
This problem is simpler if the computer only considered the relative velocity vA − vB , and their relative
position sB − sA . Then, the goal is to reduce the initial relative velocity, vA,0 − vB , to zero during
time T , while car A advances by distance D0 relative to car B. Hence, vA,0 − vB = aT , and D0 =
(vA,0 − vB )T − aT 2 /2 = (vA,0 − vB )T /2.
A different interpretation of the problem is that car A should come to rest just as it
is about to touch car B. In this view, the deceleration a of car A should be given by
a = VA,0 /T , so that vA,0 T − aT 2 /2 = vA,0 T /2 = D0 + vB T , which implies that
T = −2D0 /(2VB − VA,0 ) = −5.4 s and a = −4.5 m/s2 . What actually happens here is that
car A starts from rest at time T < 0 such that that ‘‘initial’’ conditions of the problem
are achieved at time t = 0. The negative deceleration, a = −4.5 m/s2 , means positive
acceleration. There is no solution in which car A catches up to car B such that car A
comes to rest just as it reaches car B!
QUIZ 1
Physics 101, Fall 2008
September 19, 2008
Problem 2. Romeo and Giulietta decide to exchange their Princeton class rings as tokens of
their affection (for PHY101?). Giulietta is on the balcony of her dormitory room when Romeo
tosses his ring upward with initial velocity v = 14 m/s. However, this velocity is insufficient for
the ring to reach Giulietta. Nonetheless, at the moment that Romeo’s ring reaches its highest
point, she throws her ring downwards with an initial velocity of v = 7 m/s. By a (Shakespearean?)
coincidence, Romeo catches both rings at the same moment.
a. [2 pts] How high did Romeo’s ring travel? How long was it in the air?
When Romeo’s ring is at its highest point, vf = 0 = vi − gthighest , so thighest = v/g = 14/9.8 = 1.43
s. Romeo’s ring takes the same time to fall be to Romeo as it did to reach its highest point hR , so
tRomeos ring = 2thighest = 2.86 s.
Also, vf2 = 0 = vi2 − 2gh, so hR = vi2 /2g = (14)2 /(2 · 9.8) = 10 m.
b. [2 pts] How high is Guiletta’s balcony?
Guiletta’s ring is caught by Romeo after time tfall = thighest = v/g = 1.43 s from the moment that she
threw the ring, so the height hG of her balcony is related by
hG = v tfall + gt2fall /2 = v v/g + v2 /2g = v2 /g = (14)2 /9.8 = 20 m, noting that v = v/2.
Or, just plug in: hG = 7 · 1.43 + 9.8(1.43)2/2 = 20 m.
c. [1 pt] With what minimum velocity should Romeo have thrown his ring so that Guiletta could
catch it?
For Romeo’s ring to have reached Guiletta, he should have thrown it upwards with minimum velocity
√
√
v such that v2 = 2ghG, i.e., v = 2ghG = 2 · 9.8 · 20 = 19.8 m/s = 71.3 km/hr.