The position of a particle at time tt is s(t)=t3+3t. s(t)=t3+3t. Compute the average velocity over the time interval [2,5][2,5] and estimate the instantaneous velocity at t=2.t=2. (Give your answers as whole numbers.)

(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 () is given by the quotient of the change in position and change in time at a given interval of time. i.e

= Δs / Δt

= (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;

= (s₂ - s₁) / (t₂ - t₁)

= (140 - 14) / (5 - 2)

= 126 / 3

= 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 =

V =

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

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//