A block experiences a constant acceleration of 2.4 m/s^2 from rest. What is its velocity after 4 s?

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Multiple Choice

A block experiences a constant acceleration of 2.4 m/s^2 from rest. What is its velocity after 4 s?

Explanation:
Velocity under constant acceleration builds up linearly: v = v0 + a t. Since the block starts from rest, v0 = 0. With a = 2.4 m/s^2 and t = 4 s, the velocity is v = 0 + (2.4)(4) = 9.6 m/s. So after 4 seconds the block moves at 9.6 m/s. This matches the idea that doubling the time doubles the velocity when acceleration is constant, so the other numbers would come from different times or accelerations.

Velocity under constant acceleration builds up linearly: v = v0 + a t. Since the block starts from rest, v0 = 0. With a = 2.4 m/s^2 and t = 4 s, the velocity is v = 0 + (2.4)(4) = 9.6 m/s. So after 4 seconds the block moves at 9.6 m/s. This matches the idea that doubling the time doubles the velocity when acceleration is constant, so the other numbers would come from different times or accelerations.

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