A car is driven 1200 m north at 20.0 m/s and then driven 1600 m east at 25.0 m/s
ID: 1394541 • Letter: A
Question
A car is driven 1200 m north at 20.0 m/s and then driven 1600 m east at 25.0 m/s. What are the magnitude and direction of the displacement for this trip?a. 2000 m, 36.9 A car is driven 1200 m north at 20.0 m/s and then driven 1600 m east at 25.0 m/s. What are the magnitude and direction of the displacement for this trip?
a. 2000 m, 36.9 A car is driven 1200 m north at 20.0 m/s and then driven 1600 m east at 25.0 m/s. What are the magnitude and direction of the displacement for this trip?
a. 2000 m, 36.9 2000 m, 36.9 2000 m, 36.9 a. 2000 m, 36.9
Explanation / Answer
Given Data
d1 = 1200 m north
v1 = 20.0 m/s
d2 = 1600 m east
v2 = 25.0 m/s.
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t = (1200/20) + (1600/25)
= 60 + +64
= 124 s
Because the path is a right triangle the
magnitude of displacement = sqrt(12002 + 16002)
= 2000 m
direction of displacement = tan^-1(1600/1200)
= 53.13 North of East
average velocity = displacement / time
= 2000 / 124
= 16.1 m/s
(or)
magnitude of displacement = sqrt(12002 + 16002)
= 2000 m
direction of displacement = tan^-1(1200/1600)
= 36.8 North of East
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