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

Answer:

For the interval [1, 1.0] the average velocity is $\frac{\Delta h}{\Delta t}={Undefined}$

For the interval [1, 1.01] the average velocity is $\frac{\Delta h}{\Delta t} = -8.349 \:\frac{m}{s}$

For the interval [1, 1.005] the average velocity is $\frac{\Delta h}{\Delta t} = -8.325 \:\frac{m}{s}$

For the interval [1, 1.001] the average velocity is $\frac{\Delta h}{\Delta t} = -8.305 \:\frac{m}{s}$

The instantaneous velocity at t = 1 is $-8.3 \:\frac{m}{s}$.

Explanation:

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

$\frac{\Delta s}{\Delta t} =\frac{s(t_1)-s(t_0)}{t_1-t_0}$

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 $\frac{\Delta h}{\Delta t} = \frac{(1.5t_1 - 4.9t_1^2)-(1.5t_0 - 4.9t_0^2)}{t_1-t_0}$

Consider the interval [1, 1.0]

$\frac{\Delta h}{\Delta t} = \frac{(1.5(1.0) - 4.9(1.0)^2)-(1.5(1) - 4.9(1)^2)}{1.0-1}\\\\\frac{\Delta h}{\Delta t} =\{Undefined}$

For the interval [1, 1.01]

$\frac{\Delta h}{\Delta t} = \frac{(1.5(1.01) - 4.9(1.01)^2)-(1.5(1) - 4.9(1)^2)}{1.01-1}\\\\\frac{\Delta h}{\Delta t} = -8.349 \:\frac{m}{s}$

For the interval [1, 1.005]

$\frac{\Delta h}{\Delta t} = \frac{(1.5(1.005) - 4.9(1.005)^2)-(1.5(1) - 4.9(1)^2)}{1.005-1}\\\\\frac{\Delta h}{\Delta t} = -8.325 \:\frac{m}{s}$

For the interval [1, 1.001]

$\frac{\Delta h}{\Delta t} = \frac{(1.5(1.001) - 4.9(1.001)^2)-(1.5(1) - 4.9(1)^2)}{1.001-1}\\\\\frac{\Delta h}{\Delta t} = -8.305 \:\frac{m}{s}$

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 $-8.3 \:\frac{m}{s}$. This suggests that this value is a good candidate for the instantaneous velocity at t = 1.