(a) 42m/s
(b) 15m/s
Given:
The position of a particle at time t
s(t) = t³ + 3t
(i) To compute the average velocity
Average velocity ([tex]V_{avg}[/tex]) is given by the quotient of the change in position and change in time at a given interval of time. i.e
[tex]V_{avg}[/tex] = Δs / Δt
[tex]V_{avg}[/tex] = (s₂ - s₁) / (t₂ - t₁) --------------------(ii)
Given interval of time is [2,5]
Therefore,
t₁ = 2
t₂ = 5
s₁ = position of the particle at t₁.
This is found by substituting t = 2 into equation (i)
s₁ = (2)³ + 3(2)
s₁ = 8 + 6 = 14
s₂ = position of the particle at t₂
This is found by substituting t = 5 into equation (i)
s₂ = (5)³ + 3(5)
s₂ = 125 + 15 = 140
Now, substitute t₁, t₂, s₁ and s₂ into equation (ii) as follows;
[tex]V_{avg}[/tex] = (s₂ - s₁) / (t₂ - t₁)
[tex]V_{avg}[/tex] = (140 - 14) / (5 - 2)
[tex]V_{avg}[/tex] = 126 / 3
[tex]V_{avg}[/tex] = 42
Therefore, the average velocity is 42m/s
(ii) To compute the instantaneous velocity.
The instantaneous velocity is the velocity of the particle at a given instant in time.
The given instant in time is t = 2.
To get the instantaneous velocity (V), differentiate equation (i) with respect to t as follows;
V = [tex]\frac{ds}{dt}[/tex]
V = [tex]\frac{d(t^3 + 3t)}{dt}[/tex]
V = 3t² + 3
Now substitute the value of t = 2 into the above equation
V = 3(2)² + 3
V = 12 + 3
V = 15
Therefore, the instantaneous velocity at t = 2 is 15m/s
