# 150 107 1875 direction tan 11875 619 above x axis

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15010.7= 1.875Direction: ࠵? = tan−1(1.875) = 61.9above +x axis. Magnitude: ࠵? = ࠵?࠵?sin ࠵?=210sin(61.9)= 238 ࠵?/࠵?Find x when y = 90 m. Solve for y as a function of x. Differentiate with respect to x to obtain the slope of the curve when y = 90 and x = 10.67. This gives the direction of the velocity vector at that instant. Note that ࠵?̇ = 210 =constant. Acceleration: n-t coordinates. METHOD 1. ࠵? = √150 ࠵?+ 50࠵?࠵?࠵?࠵?= (12) √150(࠵?−12) =1215010.7= 1.875࠵?2࠵?࠵?࠵?2= (−14) √150(࠵?−32)= (−√1504) (10.67−32) = −.08789METHOD 2. (࠵? − 50)2= 150࠵?2(࠵? − 50)࠵?࠵?࠵?࠵?= 150࠵?࠵?࠵?࠵?=75࠵? − 50= 1.875࠵?2࠵?࠵?࠵?2= (−1)(75)(࠵? − 50)−2(࠵?࠵?࠵?࠵?)= (−75)(40)−2(1.875) = −.08789 ࠵? =[1 + (࠵?࠵?࠵?࠵?)2]32|࠵?2࠵?࠵?࠵?2|=(1 + 1.8752)1.5. 08789= 109 ࠵?࠵?࠵?=࠵?2࠵?=2382109= 518.7 ࠵?࠵?−2࠵? =࠵?࠵?cos(90 − 61.9)= 588 ࠵?࠵?−2Direction: ࠵? = 0(directed horizontally to the right) Differentiate (twice) with respect to x to find the radius of curvature of the path when y = 90m and x = 10.67 m. Calculate the normal component of acceleration an, which is perpendicular to the direction of motion; anis directed (90 61.9°) below horizontal. Because ࠵?̇is constant, ࠵?̈ = 0therefore the total acceleration vector must be horizontal (in the +x direction).
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