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A rocket can rise to a height of h(t) = t³ + 0.3t² feet in t seconds. Find its velocity and acceleration 14 seconds after it is launched.

User Chely
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Final answer:

To answer the student's question, we calculate the first and second derivatives of the height function h(t) to find the velocity and acceleration, respectively, and evaluate them at t = 14 seconds.

Step-by-step explanation:

The student has asked for the velocity and acceleration of a rocket 14 seconds after launch, given the height function h(t) = t³ + 0.3t² feet. To find the velocity, we need to calculate the first derivative of the height function, which gives us v(t) = h'(t) = 3t² + 0.6t. Then, we evaluate this at t = 14 seconds to get the velocity at that specific time.

To find the acceleration, we take the second derivative of the height function, which yields a(t) = v'(t) = 6t + 0.6. Evaluating this at t = 14 seconds gives us the acceleration.

Now we calculate:

  • Velocity: v(14) = 3(14)² + 0.6(14)
  • Acceleration: a(14) = 6(14) + 0.6

By computing these expressions, we obtain the rocket's velocity and acceleration at 14 seconds after launch.

User Commandiron
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