Question

Use the definition of the derivative to find an expression for the instantaneous velocity of an object moving with rectilinear motion according to the given function relating s (in ft) and t (in s). Then calculate the instantaneous velocity for the given value of t. s = 8t + 19; t=4 (Simplify your answer)

Answer 1

The instantaneous velocity for
`s(t) = 8t + 19` when t = 4 is
`v(t)=8 \:(ft)/(s)`.

Explanation:

The average rate of change of function f over the interval is given by this expression:

`(f(b)-f(a))/(b-a)`

The average velocity is the average rate of change of distance with respect to time

`average \:velocity=(distance \:traveled)/(time \:elapsed) =(\Delta s)/(\Delta t)`

The instantaneous rate of change is defined to be the result of computing the average rate of change over smaller and smaller intervals.

The derivative of f with respect to x, is the instantaneous rate of change of f with respect to x and is thus given by the formula

`f'(x)= \lim_(h \to 0) (f(x+h)-f(x))/(h)`

For any equation of motion s(t), we define what we call the instantaneous velocity at time t to be the limit of the average velocity, between t and t + Δt, as Δt approaches 0.

`v(t)=\lim_(\Delta t \to 0) (s(t+\Delta t)-s(t))/(\Delta t)`

We know the equation of motion
`s(t) = 8t + 19` and we want to find the instantaneous velocity for the given value of t = 4.

Applying the definition of instantaneous velocity, we get

`v(t)=\lim_(\Delta t \to 0) (s(t+\Delta t)-s(t))/(\Delta t)\\\\v(4)=\lim_(\Delta t \to 0) (8(4+\Delta t)+19-(8(4)+19))/(\Delta t)\\\\v(t)=\lim_(\Delta t \to 0) (32+8\Delta t+19-32-19)/(\Delta t)\\\\v(t)=\lim_(\Delta t \to 0) (8\Delta t)/(\Delta t)\\\\v(t)=8 \:(ft)/(s)`

The instantaneous velocity for
`s(t) = 8t + 19` when t = 4 is
`v(t)=8 \:(ft)/(s)`.