Answer: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
