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

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

Step-by-step explanation:

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

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58