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Solve the given problems by finding the appropriate derivatives.A bullet is fired vertically upward. Its distance $s$ (in $\mathrm{ft}$ ) above the ground is given by $s=2250 t-16.1 t^{2},$ where $t$ is the time (in s). Find the acceleration of the bullet.

Calculus 1 / AB

Chapter 23

The Derivative

Section 9

Higher Derivatives

Derivatives

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In this problem, we have announced to find the acceleration of a bullet where its distance S. Is given as 2 to 50 times t minus 16.1 p square. Now the acceleration is the double derivative of the displacement S. So we need to find the double derivative which is the second derivative of S. So first of all that is determined the first derivative. So that will be 2 to 50 times the derivative of tea, which is one minus 16.1 times the derivative of T square which is two T. So we have 2 to 50 minus 32.2 T. Next we find the second derivative that will be equal to the derivative of this expression over here. So that will be the derivative of 2 to 50 which is zero since 2 to 50 is a constant minus 32.2 times the derivative of tea, which is equal to one. So this will be equal to minus 32.2. Hence the acceleration of the bullet will be given as minus 32.2, and the unit will be feet per second square, since the distance S is given in feet and p is given in seconds. Hence the acceleration is minus 32.2 ft per second square.

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