Question

During a very quick stop, a car decelerates at 7.05 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.) -26.11 Correct: Your answer is correct. rad/s2 (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 96.0 rad/s? 28 Correct: Your answer is correct. revolutions (c) How long (in s) does the car take to stop completely? 3.676 Correct: Your answer is correct. s (d) What distance (in m) does the car travel in this time? 95.11 Incorrect: Your answer is incorrect. m (e) What was the car's initial velocity (in m/s)? (Indicate the direction with the sign of your answer.)

Answer 1

Part a)

Initial deceleration is given as

$a = - 7.05 m/s^2$

radius is given as

$R = 0.270 m$

now angular acceleration is given as

$\alpha = \frac{a}{R}$

$\alpha = \frac{-7.05}{0.270} = -26.11 rad/s^2$

Part b)

initial angular speed = 96 rad/s

angular deceleration = 7.05 rad/s

now number of revolutions before it will stop is given as

$N = \frac{\omega^2 - \omega_0^2}{4\pi\alpha}$

now plug in all data

$N = \frac{0 - 96^2}{4\pi(-26.11)}$

$N = 28 rev$

Part c)

time taken to stop is given as t

now we have

$\omega = \omega_0 + \alpha t$

$0 = 96 - 26.11 t$

$t = 3.676 s$

Part d)

distance moved by the car

distance = number of revolutions (length of one complete revolution)

$d = N(2\pi R)$

$d = 28(2\pi (0.270))$

$d = 47.5 m$

Part e)

initial speed of car is given as

$v = R\omega$

$v = 0.270(96) m/s$

$v = 25.92 m/s$