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A ball is thrown vertically from the top of a building. The height of the ball after t seconds can be given by the function s(t)= -0.1(t-2)^2 + 10
meters. What is the estimated instantaneous velocity of the ball after 4 seconds.

Respuesta :

Ashraf82

The instantaneous velocity of the ball after 4 seconds is -0.4 m/s

Step-by-step explanation:

If f(x) is the function which represents the distance that a particle moves after x seconds, with velocity v and acceleration a, then

  • v(x) = f'(x) ⇒ first derivative
  • a(x) = f"(x) ⇒ second derivative

∾ A ball is thrown vertically from the top of a building

∾ The height of the ball after t seconds can be given by the

   function s(t)= -0.1(t -2)² + 10

- To find the function of the velocity differentiate s(t)

∾ s(t) = -0.1(t - 2)² + 10

- Solve the bracket

∾ (t - 2)² = t² - 4t + 4

∴ s(t) = -0.1(t² - 4t + 4) + 10

- Multiply the bracket by -0.1

∴ s(t) = -0.1t² + 0.4t - 0.4 + 10

- Add the like terms

∴ s(t) = -0.1 t² + 0.4t + 9.6

Now let us differentiate s(t)

∾ s'(t) = -0.1(2)t + 0.4(1)

∴ s'(t) = -0.2t + 0.4

- s'(t) is the function of velocity after time t seconds

∾ s'(t) = v(t)

∴ v(t) = -0.2t + 0.4

We need to find the instantaneous velocity of the ball after 4 seconds

Substitute t by 4

∴ v(4) = -0.2(4) + 0.4

∴ v(4) = -0.8 + 0.4

∴ v(4) = -0.4

∴ The v is -0.4 m/s ⇒ -ve means the velocity is downward

The instantaneous velocity of the ball after 4 seconds is -0.4 m/s

Learn more:

You can learn more about the instantaneous velocity in brainly.com/question/2234298

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