Awheel of radius 0.23 m, which is moving initially at 25.0 m/s, rolls to a stop in 246.0 m. the wheel's rotational inertia is 0.170 kg·m2 about its central axis. calculate the magnitude of its linear acceleration.

-1.27 m/s^2

Explanation:

a = - V^2 / 2x

a = -(25^2) / 2 x (246) = 1.27 m/ s^2

Therefore the linear acceleration of the wheel is - 1.27 m/s^2

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