Question

During a very quick stop, a car decelerates at 6.2 m/s2. Assume the forward motion of the car corresponds to a positive direction for the rotation of the tires (and that they do not slip on the pavement). Randomized Variables at = 6.2 m/s2 r = 0.275 m ω0 = 93 rad/s
Part (a) What is the angular acceleration of its tires in rad/s2, assuming they have a radius of 0.275 m and do not slip on the pavement? α =

Part (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 93 rad/s ?

Part (c) How long does the car take to stop completely in seconds?

Part (d) What distance does the car travel in this time in meters?

Part (e) What was the car’s initial speed in m/s?

Answer 1

Answer:

a) $\alpha = -22.545\,\frac{rad}{s^{2}}$, b) $n = 30.528\,rev$, c) $t = 4.125\,s$, d) $s = 52.748\,m$, e) $v_{o} = 25.575\,\frac{m}{s}$

Explanation:

a) A constant deceleration means a constant linear deceleration. The angular acceleration experimented by the tires is:

$\alpha = \frac{-6.2\,\frac{rad}{s^{2}} }{0.275\,m}$

$\alpha = -22.545\,\frac{rad}{s^{2}}$

b) The period of time required to decelerate the car is derive from the following expression:

$t = \frac{\omega - \omega_{o}}{\alpha}$

$t = \frac{0\,\frac{rad}{s}- 93\,\frac{rad}{s} }{-22.545\,\frac{rad}{s^{2}} }$

$t = 4.125\,s$

The angular position at given time is:

$\theta = (93\,\frac{rad}{s} )\cdot (4.125\,s)+\frac{1}{2}\cdot (-22.545\,\frac{rad}{s^{2}} )\cdot (4.125\,s)^{2}$

$\theta = 191.816\,rad$

The number of revolutions are:

$n =\frac{\theta}{2\pi}$

$n = 30.528\,rev$

c) The time needed to decelerate the car is:

$t = 4.125\,s$

d) The initial speed experimented by the car is:

$v_{o} = (93\,\frac{rad}{s} )\cdot (0.275\,m)$

$v_{o} = 25.575\,\frac{m}{s}$

The distance done by the car during its deceleration is:

$s = (25.575\,\frac{m}{s})\cdot (4.125\,s)+\frac{1}{2}\cdot (-6.2\,\frac{m}{s^{2}} )\cdot (4.125\,s)^{2}$

$s = 52.748\,m$

e) The initial speed of the car is:

$v_{o} = 25.575\,\frac{m}{s}$