The car is traveling at a speed of 60 mi/hr as it approaches point A. Beginning

Question

The car is traveling at a speed of 60 mi/hr as it approaches point A. Beginning at A, the car decelerates at a constant 7 ft/sec2 until it gets to point B. after which its constant rate of decrease of speed is 3 ft/sec2 as it rounds the interchange ramp. Determine the magnitude of the total car acceleration (a) just before it gets to B, (b) just after it passes B, and (c) at point C.

Explanation / Answer

velocity at A : Va = 60mil/hr = 60*5280/3600 ft/sec = 88ft/sec velocity at B : Vb (Va)^2 -(Vb)^2 = 2*a*s 7744 - (Vb)^2 = 2*(7)*300 ft^2/sec^2 (Vb)^2 = 3544ft^2/s^2 Vb = 59.53ft/s (Vb)^2 - (Vc)^2 = 2*3 * (3.14*200/2) 3544 - (Vc)^2= 1884ft^2/s^2 Vc= 40.74ft/s tangential acceleration at C: At = (Vc)^2/R = 8.3 ft/s^2 Total acceleration at C : Ac (Ac)^2 = (3)^2 + (8.3)^2 Ac = 8.82ft/s2 a) Just before point B: Accelaration -> a = -7ft/s2 b)Just after point B: Accelaration -> a = -3ft/s2 c) At C Acceleration : Ac = *.82ft/s ( or 8.3i - 3j)ft/s^2

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