University Physics
Quiz 3 Solutions
Fri Sep 14, 2012
Your name:
Please show your work, write neatly, and put a box or circle around each of your answers.
This week the quiz is also being given in MasteringPhysics with the same problems.
1. (5 points total) In a chase scene, a movie stuntman runs horizontally off a flat roof of
one building and lands on another roof 2.0 m lower.
(a) If the gap between the buildings is 4.9 m wide, how fast must he run to just barely
cross the gap? In other words, what is his initial horizontal velocity vx0 (he’s
running horizontally) so he barely crosses the horizontal distance between the two
buildings? Express your answer to two significant figures and include the appropriate units.
Solution: Just like the “dropping the bomb” example in class, here you know
that the initial vertical velocity is zero (vy0 = 0 m/s), and the distance the stuntman is dropping due to gravity. You can then use the vertical distance equation to
solve for time:
s
s
1
1
2(y − y0 )
2(−2.0 m)
y − y0 = vy0 t + at2 = gt2 ⇒ t =
=
= 0.64 s.
2
2
g
−9.8 m/s2
The stuntman has to cover the horizontal distance x − x0 = 4.9 m in this time, so
we can calculate his initial horizontal velocity:
x − x0 = vx0 t
⇒
vx0 =
(x − x0 )
(4.9 m)
=
= 7.7 m/s = vx0
t
0.64 s
(b) If the stuntman runs twice as fast horizontally off the roof, how much of a horizontal
gap can he cover?
i. Four times the original gap.
ii. Twice the original gap. since he still falls in the same amount of time but he’s
running twice as fast.
iii. The same as the original gap.
iv. Half the original gap.
v. There’s not enough information to figure out.
2. (5 points total)
(a) How fast (at constant speed) would a car have to go around a circular turn of radius
98 m for its centripetal acceleration to have the same magnitude as gravitational
acceleration, 9.8 m/s2 ? Express your answer to two significant figures and include
the appropriate units of m/s.
Solution: The centripetal acceleration equation tells us a = v 2 /r. Here we know
the desired acceleration, and the radius, so we can figure out the required speed v:
q
√
v2
acentrip =
⇒ v = ar = (9.8 m/s2 )(95 m) = 31 m/s = v
r
1
(b) Is the car’s velocity constant when going around the curve?
i. Yes.
ii. No. The speed is constant but the direction of the velocity is always changing,
so the velocity is not constant and the car is indeed accelerating.
iii. There’s not enough information provided to decide.
(c) The car comes through the same curve but with half of the speed that you calculated
in the first part. What is its acceleration in this case?
i.
ii.
iii.
iv.
a=g/2 pointed inwards towards the center of the curve.
a=0 since the speed is constant.
a=g/4 pointed outwards from the center of the curve.
a=g/2 pointed outwards from the center of the curve.
v. a = g/4 pointed inwards towards the center of the curve. The acceleration is
proportional to the speed squared, so one half of the speed requires one fourth
of the centripetal acceleration to go around the same curve.
2