Answer to Question #177166 in Classical Mechanics for Joseph Se

Question

The final answer should be in the format of "tolerance of ± in the third significant digit."

The car is traveling at a speed of 69 mi/hr as it approaches point A. Beginning at A, the car decelerates at a constant 6.9 ft/sec² until it gets to point B, after which its constant rate of decrease of speed is 2.0 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.

Distance from A to B is 325’

Radius is 243’

(a) Just before B,a = ________ ft/sec²

(b) Just after B,a = ______ ft/sec²

(c) At C,a = _______  ft/sec²

Answer 1

(a) Just before B, a = 6.9 ft/sec²

(b) Just after B,a = 2 ft/sec²

(c) At C, after one fourth of the circle, the tangential acceleration is 2 ft/sec². The centripetal acceleration depends on speed. So, find the speed at B:

$$v_B=\sqrt{v_0^2-2a_1d_{AB}}=51.1\text{ mi/hr}.$$

Then the velocity continues to decrease at 2 ft/sec²:

$$v_C=\sqrt{v_B^2-2a_2d_{BC}}=\sqrt{v_B^2-2a_2·(\pi R/2)}=\\\space\\=\sqrt{v_B^2-\pi a_2R}=44.3\text{ mi/hr}.$$

$$a_c=\frac{v_B^2}{R}=17.4\text{ ft/s}^2.\\\space\\ a=\sqrt{a_\tau^2+a_c^2}=17.5\text{ ft/sec}^2.$$