during a very quick stop, a car decelerates at 6.75 m/s2. (a) what is the angular acceleration (in rad/s2) of its 0.270 m radius tires, assuming they do not slip on the pavement? (indicate the direction with the sign of your answer. assume the tires initially rotated in the positive direction.) rad/s2 (b) how many revolutions do the tires make before coming to rest, given their initial angular velocity is 94.0 rad/s? revolutions (c) how long (in s) does the car take to stop completely? s (d) what distance (in m) does the car travel in this time? m (e) what was the car's initial velocity (in m/s)? (indicate the direction with the sign of your answer.) m/s (f) do the values obtained seem reasonable, considering that this stop happens very quickly?



Answer :

a) Angular acceleration (in rad/s2) = 6.75 m/s2 * (1 m / 0.270 m) = 25 rad/s2 (in the negative direction, since the car is decelerating)

b) Number of revolutions = (94.0 rad/s) / (25 rad/s2) = 3.76 revolutions

c) Time to stop completely = (94.0 rad/s) / (25 rad/s2) = 3.76 s

d) Distance travelled = (94.0 rad/s * 3.76 s) + (1/2 * 25 rad/s2 * 3.76 s2) = 402.5 m

e) Initial velocity = (25 rad/s2 * 3.76 s) + (94.0 rad/s) = 117 m/s (in the negative direction, since the car is decelerating)

f) The values obtained seem reasonable, considering that this stop happens very quickly. The initial velocity of 117 m/s is quite large, indicating that the car was traveling at a significant speed prior to the stop. The distance traveled, 402.5 m, is also reasonable.

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