A wheel originally rotating at 126 rad/s undergo a constant angular deceleration of 5.00rad/s. What is the angular speed after it has turned through an angel of 628 radians?

98.0 rad/s The deceleration in this question is improperly described at 5.00 rad/s, instead of the correct value of 5.00 rad/s^2. But that's easily corrected and the formula for the number of rotations taken then becomes R = VT - 0.5AT^2 where R = number of rotations V = Initial velocity T = Time A = Acceleration. So let's see how much time it takes for perform 628 radians of rotation. R = VT - 0.5AT^2 628 = 126T - 0.5 * 5.0 T^2 628 = 126T - 2.5 T^2 0 = 126T - 2.5 T^2 - 628 We now have a quadratic equation for A=-2.5, B=126, and C=-628. Use the quadratic formula to get the roots of 5.608164966 and 44.79183503 seconds. The lower value is the one we want. The larger value is how many seconds it would take for the wheel to come to a stop and continue to accelerate in the reverse direction until it back tracked far enough to get back to the desired 628 radians. Since we now know that it takes 5.608164966 seconds to get the desired number of rotations, we can easily figure out the wheel's velocity at that moment by subtracting from the original velocity the acceleration multiplied by the time. So: 126 rad/s - 5 rad/s^2 * 5.608164966 s = 126 rad/s - 28.04082483 rad/s = 97.95917517 rad/s Rounding to 3 significant figures gives 98.0 rad/s