In the overhead view of Figure 4-47, Jeeps P and B race along straight lines, ac
ID: 2020510 • Letter: I
Question
In the overhead view of Figure 4-47, 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 18.0 m/s, at the angle theta2 = 34.0degree. Relative to the guard, P has accelerated from rest at a constant rate of 0.460 m/s2 at the angle theta1 = 63.0degree. At a certain time during the acceleration, P has a speed of 43.0 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
The velocity of Jeep P relative to A at the instant is (in m/s) VpA = 40.0(cos60 ˆi + sin 60 ˆj) = 20.0ˆi +34.6ˆj. Similarly, the velocity of Jeep B relative to A at the instant is (in m/s) VBA = 20.0(cos30 ˆi + sin 30 ˆj) = 17.3ˆi + 10.0ˆj. Thus, the velocity of P relative to B is (in m/s) VPB = VPA-VBA = (20.0ˆi + 34.6ˆj) - (17.3ˆi + 10.0ˆj) = 2.68ˆi+24.6ˆj ........................................................................................... (a) The magnitude of VPB = sqrt [(2.68)2+(24.6)2] = 24.8 m/sec (b) The direction of VPB = = tan-1 (24.6 / 2.68) = 83.80 north of east (or 6.2º east of north). ................................................................................................................ (c) The acceleration of P is aPA = 0.400 (cos 60.0 ˆi + sin 60.0 ˆj) = 0.200ˆi + 0.346ˆj and aPA = aPB Thus, we have aPB = 0.400 m/s2. (d) The direction is 60.0° north of east (or 30.0° east of north).
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