TA: Tomoyuki Nakayama
Monday, September 20, 2010
PHY 2048: Physic 1, Discussion Section 7193
Quiz 3 (Homework Set #4)
Name:
UFID:
Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator.
You need to show all of your work for full credit.
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In the overhead view of the figure below right, Jeeps P and B race along straight lines, across flat
terrain, and past stationary border guard A. Relative to the guard, P travels at a constant speed of
30.0 m/s, at the angle θ1 = 65.0º. Relative to the guard, B has accelerated from rest at a constant rate
of 0.500 m/s2 at the angle θ2 = 35.0º. At a certain time during the acceleration, B has a speed of 20.0
m/s.
a) At that time, what are the magnitude and
direction of the velocity of P relative to B?
We take our +x axis to the east and +y axis to
the north. In unit-vector notation, the velocity of
P with respect to stationary guard A is
vPA = 30cosθ1i + 30sinθ1j = 12.7i + 27.2j
The velocity of B with respect to A is
vBA = 20cosθ2i + 20sinθ2j = 16.4i + 11.5j
The velocity of P relative to B is
vPB = vPA – vBA = -3.7i + 15.7j
The magnitude of vPB is
vPB = √((-3.7)2+15.72) = 16.1 m/s.
The direction is
θ = tan-1(15.7/(-3.7)) = -76.7 + 180 = 103º, 13º west of north.
Note that vPB has a negative x component and positive y component, so it is in the 2nd quadrant.
b) At that time, what are the magnitude and direction of the acceleration of B relative to P?
Since P moves at a constant velocity, its acceleration is zero. Thus we have
aBP = aBA – aPA = aBA
It has a magnitude of 0.500 m/s2 and is directed 35º north of east.