High School

If a ball is thrown into the air from ground level with an initial velocity of 122 ft/sec, its height (in feet) after \( t \) seconds is given by the position function:

\[ s(t) = 122t - 16t^2 \]

Find the velocity (in ft/sec) of the ball after \( t \) seconds.

A. \( 122 - 32t \)
B. \( 122 + 32t \)
C. \( 122 - 16t \)
D. \( 122 + 16t \)

Answer :

Final answer:

To find the ball's velocity after t seconds, differentiate the position function s(t) = 122t - 16t^2 to obtain the velocity function v(t) = 122 - 32t, indicating answer option a) 122 - 32t is correct.

Explanation:

To find the velocity (in ft/sec) of the ball after t seconds, we need to differentiate the position function s(t) = 122t - 16t^2 with respect to t. The differentiation of s(t) will give us the velocity function v(t), which represents the rate of change of position, or in other words, the velocity of the ball at any time t.

Differentiating s(t) yields:

  • The derivative of 122t is 122.
  • The derivative of -16t^2 is -32t, by applying the power rule (multiply the exponent by the coefficient and decrease the exponent by one).

Therefore, the velocity function v(t) after t seconds is 122 - 32t.

This means the correct answer is a) 122 - 32t, representing the velocity of the ball after t seconds in feet per second.

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