I need some help in how to solve this dynamics problem. I thought I did the prob
ID: 1816778 • Letter: I
Question
I need some help in how to solve this dynamics problem. I thought I did the problem right but when I checked my answer it was wrong. And im stuck on what im doing wrong. Here is the description of the problem:A motorist traveling along a straight portion of a highway is decreasing the speed of his automobile at a constant rate before exiting from the highway onto a circular exit ramp with a radius of 560 feet. He continues to decelerate at the same constant rate so that 10 seconds after entering the ramp, his speed has decreased to 20 miles per hour, a speed which he then maintains. Knowing that at this constant speed the total acceleration of the automobile is equal to one-quarter of its value prior to entering the ramp, determine the maximum value of the total acceleration of the automobile.
Explanation / Answer
At his constant speed of 20 mph his acceleration is purely centripetal acceleration, which is a = v^2/r. 20 mph is equivalent to 29.333 feet/second (I won't post how to do it, its easy). 29.333 ft/s ^2 / 560 feet = 1.536 ft/s^2 acceleration. This is one-quarter of his acceleration PRIOR to getting on the ramp. 1.536 * 4 = 6.144 ft/s^2 before getting on the ramp. When he gets on the ramp, his acceleration is both due to braking and being on the ramp. We need his speed when he first gets on the ramp for the maximum acceleration of the automobile. v = v0 + a*t (basic physics equation). We have v, a and t, we are looking for v0. v0 = v - a*t = 29.333 - (-6.144*10) = 90.773 feet/second. centripetal acceleration = v^2/r = 90.773 ^2 / 560 = 14.714 feet/sec^2 As both accelerations are perpendicular to each other we can't simply add them, instead we have to use pythagoreans theorem. a^2 = b^2 + c^2 = 6.144^2 + 14.714^2 a^2 = 254.250 a = 15.945 feet/second^2