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 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.

CLWOMY The Asker · Mechanical Engineering

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.

Transcribed Image Text: Ee 2/111 The car is traveling at a speed of 60 mi/hr as it ap- proaches point A. Beginning at A, the car deceler- ates at a constant 7 ft/sec? until it gets to point B, after which its constant rate of decrease of speed is 3 ft/sec? as it rounds the interchange ramp. Deter- mine 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. Problem 2/111
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Transcribed Image Text: Ee 2/111 The car is traveling at a speed of 60 mi/hr as it ap- proaches point A. Beginning at A, the car deceler- ates at a constant 7 ft/sec? until it gets to point B, after which its constant rate of decrease of speed is 3 ft/sec? as it rounds the interchange ramp. Deter- mine 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. Problem 2/111
Community Answer
INEOW8

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