During a panic stop, a car decelerates at 7.00 m/s^2. (a) What is the angular ac

Question

During a panic stop, a car decelerates at 7.00 m/s^2. (a) What is the angular acceleration of it's 0.280 m radius tires, assuming they do not slip on the pavement? (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 95.0 rad/s? (c) How long does it take? (d) What distance does the car travel in this time? (e) What was its initial velocity? (f) Do the values obtained seem reasonable, considering that this is a panic stop?

The answers are (a) -25.0 rad/s^2 (b) 28.7 rev (c) 3.80 s (d) 50.7 m (e) 26.6 m/s.

I need to know the steps you need to take to find out those answers.

Explanation / Answer

a). for no-slip, linear acceleration = angular acceleration*radius => -7 = angular acceleration*0.28 => angular acceleration = -25 rad/s^2 b). wf^2 - wi^2 = 2*angular acceleration*theta => 0 - 95^2 = 2*(-25)*theta => theta = 180.5 radian = 180.5/(2*pi) rev = 28.74 rev c). w = w0 + angular acceleration*t => 0 = 95 +(-25)*t => t = 3.8 s d). distance travelled = no. of revolutions*(2*pi*r) = 28.74*2*3.14*0.28 = 50.54 m e). initial velocity = initial angular velocity*radius = 95*0.28 = 26.6 m/s f). No, because distance travelled before coming to stop(i.e., 50.54 m) is very high for a panic stop.

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