Question

If an object&#x27;s acceleration is defined as a time-varying function a(t) = 6t, what is the exact mathematical expression for its velocity after 4 seconds, assuming it started from rest?

Accepted Answer

To find velocity from acceleration, you must perform integration, a mathematical process that finds the total accumulation of a changing rate over time. Since acceleration is the derivative of velocity—meaning it represents how fast velocity is changing—we reverse this process by integrating the function a(t) = 6t with respect to time. The integral of 6t is 3t², which gives us the general velocity function v(t) = 3t² + C. The constant C represents the initial velocity, or the velocity at time zero; because the problem states the object started from rest, C is 0, leaving us with the specific velocity function v(t) = 3t². To find the velocity at 4 seconds, we substitute 4 for t, resulting in 3(4²) = 3(16) = 48. Thinking of this like filling a bucket with water where the flow rate increases over time, the integration accounts for every drop added until you reach the total volume at a specific moment. Therefore, the velocity after 4 seconds is exactly 48 units of velocity per unit of time.

If an object's acceleration is defined as a time-varying function a(t) = 6t, what is the exact mathematical expression for its velocity after 4 seconds, assuming it started from rest?

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If an object's acceleration is defined as a time-varying function a(t) = 6t, what is the exact mathematical expression for its velocity after 4 seconds, assuming it started from rest?

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