A car initially traveling eastward turns north by traveling in a circular path a

Question

A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in the figure below. The length of the arc ABC is 237 m, and the car completes the turn in 32.0 s. (a) What is the acceleration when the car is at B located at an angle of 35.0degree? Express your answer in terms of the unit vectors i and j. First find the magnitude of the acceleration; then work out the components of the vector. m/s^2 i + (b) Determine the car's average speed. m/s (c) Determine its average acceleration during the 32.0-s interval. m/s^2 i + m/s^2 j

Explanation / Answer

arc length of ABC =237m

time taken t=32 sec

a)

radius of the path, r=l/theta

r=237/(pi/2)

r=150.88 m

and

speed v=l/t

=237/32

=7.406 m/sec

and

magnitude of acceleration a=v^2/r

a=7.406^2/(150.88)

a=0.36 m/sec^2 ----------------

in vector notation,

a=(ax)i+(ay)j

a=(a*cos(theta))i+(a*sin(theta))j

a=(0.36*cos(360-35))i+(0.36*sin(360-35))j

a=(0.29)i-(0.21)j

b)

avg speed v=l/t

=237/32

=7.406 m/sec

c)

avg acceleraion, a=(0.29)i-(0.21)j

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