A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:
A) 17 s B) 15 s C) 19 s D) 21 s E) 23 s

Question

A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:
A) 17 s B) 15 s C) 19 s D) 21 s E) 23 s

anonymous · Answer

it really close to 23s in not really sure but i did calculation in my head and i trust it right
lol

anonymous · Answer

can you provide your steps, thanks

anonymous · Answer

$$\alpha = \frac{ d \omega }{ dt }$$Angular acceleration is derivative of angular velocity  the same way as linear acceleration is derivative of linear velocity $$a = \frac{ dv }{ dt}$$

anonymous · Answer

To make I'm on the same page with you
I took these steps

t = (25 rad/s) / 1.5 rad/s^2 = 16.6 ~ 17s