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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

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 265 m, and the car completes the turn in 36.0 s.

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

a) What is the acceleration when the car is at B located at an angle of 35.0°? Express your answer in terms of the unit vectors i and j.

(m/s2 i + m/s2 j)

(c) Determine its average acceleration during the 36.0 s interval.

(m/s2 i m/s2 j)

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Answers

Hi Jerica, first we make the calculation:

Time and length is given. V = 365 m / 36 s = -----m/s

average acceleration is a=(2 * V) / t = -----m/s2, it is the radial vector

then we have the x-vector in 35° = cos(35) * a = ----- m/s2

then we have the y-vector in 35° = sin(35) * a = ----- m/s2

I hope you can translate x / y to i / j. To understand the whole calculation, you have to watch the whole turn with an equal speed. Right after A you have an acceleration rectangular to the drive-direction. Cos is zero and goes with the turn to 1.00. Sin goes from 1 to zero. During the turn, sin^2 + cos^2 is always 1. At 35° it will be x=cos(35°) and y=sin(35°) of the unit circle with radius of 1, what causes you to multiply each time with the acceleration. The acceleration is caused by the fact, that the speed-vector at A has to slow down of the length of 265m to zero in 36 sec.

 

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