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A toy rocket fired straight up into the air has height s(t) = 192t - 16t² feet after t seconds. a. What is the rocket's initial velocity (when't = 0)? b. What is the velocity after 4 seconds? c. What is the acceleration when t = 9? d. At what time will the rocket hit the ground? e. At what velocity will the rocket be traveling just as it smashes into the ground? ... a. The rocket's initial velocity is ft/sec

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Answer:

Explanation:

a. The initial velocity of the rocket can be determined by finding the derivative of the height function with respect to time and evaluating it at t = 0.

Given the height function: s(t) = 192t - 16t²

To find the initial velocity, take the derivative of s(t) with respect to t:

s'(t) = d/dt (192t - 16t²)

s'(t) = 192 - 32t

Now, plug in t = 0:

s'(0) = 192 - 32 * 0

s'(0) = 192 ft/sec

So, the rocket's initial velocity is 192 ft/sec.

b. To find the velocity after 4 seconds, plug in t = 4 into the derivative of the height function:

s'(t) = 192 - 32t

s'(4) = 192 - 32 * 4

s'(4) = 192 - 128

s'(4) = 64 ft/sec

The velocity after 4 seconds is 64 ft/sec.

c. The acceleration can be determined by taking the second derivative of the height function with respect to time:

s''(t) = d²/dt² (192t - 16t²)

s''(t) = -32

The acceleration is constant and is -32 ft/sec².

d. To find the time when the rocket hits the ground, set the height function s(t) equal to zero (since the rocket hits the ground when its height is zero) and solve for t:

192t - 16t² = 0

16t(12 - t) = 0

This gives two possible solutions: t = 0 and t = 12. Since we're interested in the time the rocket hits the ground, the valid solution is t = 12 seconds.

e. To find the velocity of the rocket when it hits the ground, plug in t = 12 into the derivative of the height function:

s'(t) = 192 - 32t

s'(12) = 192 - 32 * 12

s'(12) = 192 - 384

s'(12) = -192 ft/sec

The velocity when the rocket hits the ground is -192 ft/sec. The negative sign indicates that the rocket is moving downward.

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