Starting at the origin, you travel a distance 5.0 m in a direction 60.8 degrees

Question

Starting at the origin, you travel a distance 5.0 m in a direction 60.8 degrees north of east. Then, from this new postition, you travel another distance 2.6 m in a direction 41.2 degrees north of east.

a) In your final postition, how far are you from the origin? (answer in m)

b) In your final position, how many degrees north of east are you as measured from the origin? (answer in degrees)

c) Suppose that during both phases of your motion, you moved with a constant speed of 9.6 m/s. How much time does the whole trip take, from the origin to your final position? (answer in s)

d) What is the magnitude of your average velocity for the whole trip from the origin to your final position? (answer in m/s)

e) What was the magnitude of your average acceleration for the whole trip from the origin to your final position? Answer in m/s^2)

Explanation / Answer

here,

first trip , A = 5m * ( cos(60.8) i + sin(60.8) j)

A = 2.43 i + 4.36 j m

seccond trip , B = 2.6* ( cos(41.2) i + sin(41.2) j

B = 1.95 m i + 1.70 m j

total distance , C = A + B

C = 4.38 i m + 6.06 m j

a)

|C| = sqrt(4.38^2 + 6.06^2)

|C| = 7.47 m

the distance from the origin is 7.47m

b)

theta = arctan(6.06/4.38)

theta = 54.7degree

the final position is 54.7 degree north of east

c)

constant speed , v = 9.6m/s

total time taken , t = ( 5 + 2.6) / 9.6

t = 0.79s

the time taken for the whole trip is 0.79 s

d)

average velocity , v = net displacement / time taken

v = 7.47 / 0.79

v =9.45 m/s

the average velocity is 9.45 m/s

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