Asked by t

During a very quick stop, a car decelerates at 7.25 m/s2.
(a)What is the angular acceleration (in rad/s2) of its 0.250 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.)
(b)How many revolutions do the tires make before coming to rest, given their initial angular velocity is 96.0 rad/s?
(c)How long (in s) does the car take to stop completely?
(d)What distance (in m) does the car travel in this time?
(e)What was the car's initial velocity (in m/s)? (Indicate the direction with the sign of your answer.)
(f) Do the values obtained seem reasonable, considering that this stop happens very quickly?

All Answers 1

Answered by GPT-5 mini AI
Take positive as the initial direction of rotation/velocity.

(a) alpha = a/r = -7.25 / 0.250 = -29.0 rad/s^2 (negative because the tires are slowing).

(b) Use 0 = omega0^2 + 2 alpha theta → theta = -omega0^2/(2 alpha) = (96.0^2)/(2·29.0) = 158.9 rad = 158.9/(2π) = 25.3 revolutions.

(c) t = -omega0/alpha = 96.0/29.0 = 3.31 s.

(d) s = r·theta = 0.250·158.9 = 39.7 m.

(e) v0 = r·omega0 = 0.250·96.0 = +24.0 m/s (positive direction).

(f) These are reasonable: a = 7.25 m/s^2 ≈ 0.74 g (a strong but plausible braking), stopping from 24.0 m/s (≈86 km/h) in ≈3.3 s and ≈40 m is fast but realistic.