In the overhead view of the figure, Jeeps P and B race along straight lines, acr

Question

In the overhead view of the figure, Jeeps P and B race along straight lines, across flat terrain, and past stationary border guard A. Relative to the guard, B travels at a constant speed of 16.0 m/s, at the angle theta2 = 33.0 degree. Relative to the guard, P has accelerated from rest at a constant rate of 0.480 m/s^2 at the angle theta 1 = 59.0degree. At a certain time during the acceleration, P has a speed of 41.0 m/s. At that time, what are the (a) magnitude and (b) direction of the velocity of P relative to B and the (c) magnitude and (d) direction of the acceleration of P relative to B? Give directions as a positive (counterclockwise) angle relative to due east.

Explanation / Answer

We take our +x axis to the east and +y axis tothe north. In unit-vector notation, the velocity of
B with respect to stationary guard A is
vBA = 16cos1i + 16sin1j = 13.42i +8.71j
The velocity of P with respect to A is
vPA = 41cos2i + 41sin2j = 21.12i + 35.14j
The velocity of B relative to P is
vBP = vBA – vPA = -7.7i -26.43j
The magnitude of vBP is
vBP = sqrt((-7.7)2+(-26.43)2) = 27.52 m/s.
The direction is
= tan-1(3.7/(-15.7)) = 73.75 north of east.
Note that vBP has a negative x component and negative y component, so it is in the 3rd quadrant

b)
Since B moves at a constant velocity, its acceleration is zero. Thus we have
aPB = aPA – aPB = aPA
It has a magnitude of 0.480 m/s2 and is directed 59º north of east.

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