Question

A stone is tossed into the air from the ground level with an initial velocity 15 m/sec. it’s height in meters a t seconds is h(t) = 1.5T - 4.9t^2. Compute the average velocity of the stone over the given time intervals, then use that information to calculate the instantaneous velocity of the stone at t = 1sec. Show your work time intervals:

a) [1, 1.0]

b) [1, 1.01]

c) [1, 1.005]

d) [1, 1.001]

Answer 1

For the interval [1, 1.0] the average velocity is

For the interval [1, 1.01] the average velocity is
`(\Delta h)/(\Delta t) = -8.349 \:(m)/(s)`

For the interval [1, 1.005] the average velocity is
`(\Delta h)/(\Delta t) = -8.325 \:(m)/(s)`

For the interval [1, 1.001] the average velocity is
`(\Delta h)/(\Delta t) = -8.305 \:(m)/(s)`

The instantaneous velocity at t = 1 is
`-8.3 \:(m)/(s)`.

Step-by-step explanation:

The average velocity over
`[t_0,t_1]` is

`(\Delta s)/(\Delta t) =(s(t_1)-s(t_0))/(t_1-t_0)`

where

`\Delta s=s(t_1)-s(t_0)` = change in position

`\Delta t =t_1-t_0` = change in time (length of interval)

We know that the height in meters a t seconds is given

`h(t) = 1.5t - 4.9t^2`

so the average velocity is
`(\Delta h)/(\Delta t) = ((1.5t_1 - 4.9t_1^2)-(1.5t_0 - 4.9t_0^2))/(t_1-t_0)`

Consider the interval [1, 1.0]

For the interval [1, 1.01]

`(\Delta h)/(\Delta t) = ((1.5(1.01) - 4.9(1.01)^2)-(1.5(1) - 4.9(1)^2))/(1.01-1)\\\\(\Delta h)/(\Delta t) = -8.349 \:(m)/(s)`

For the interval [1, 1.005]

`(\Delta h)/(\Delta t) = ((1.5(1.005) - 4.9(1.005)^2)-(1.5(1) - 4.9(1)^2))/(1.005-1)\\\\(\Delta h)/(\Delta t) = -8.325 \:(m)/(s)`

For the interval [1, 1.001]

`(\Delta h)/(\Delta t) = ((1.5(1.001) - 4.9(1.001)^2)-(1.5(1) - 4.9(1)^2))/(1.001-1)\\\\(\Delta h)/(\Delta t) = -8.305 \:(m)/(s)`

The instantaneous rate of change is the limit of the average rates of change. We estimate the instantaneous rate of change at
`x=x_0` by computing the average rate of change over smaller and smaller intervals

From the calculations above we can see when the time interval shrinks the average velocity tends to the value
`-8.3 \:(m)/(s)`. This suggests that this value is a good candidate for the instantaneous velocity at t = 1.