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 7ft/sec^2 until it gets to point B, after which its constant rate of decrease of speed is 3 ft/sec^2 as it rounds the interchange ramp. Determine the magnitue of the total car acceleration a) just before it gets to B, b) just after it passes B, and c) at point C.

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Draw the free body diagram of the curve.(a)Write the equation for the normal and tangential acceleration from point A to B, where the curve is a straight line.{:[a_(n)=0],[a_(t)=-7ft//sec^(2)]:}Calculate the total acceleration before point B using the formula below:{:[a=sqrt((a_(t)^(2)+a_(n)^(2)))],[=sqrt(0^(2)+(-7)^(2))],[=7ft//sec^(2)]:}(b)Acceleration after the point B consists of both normal and tangential components.{:[v_(s)^(2)=v_(A)^(2)+2a_(t)(s_(s)-s_(A))],[v_(s)^(2)=88^(2)+(2xx-7xx(300-0))],[v_(s)=59.53ft//s],[a ... See the full answer