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