Question

If a ball is thrown straight up with a velocity of$40\text{ft/s,}$ its height (in ft) aftert seconds is given by$y=40t-16t^2$ .Find the instantaneous velocity when$t=2$ .

Short answer

The instantaneous velocity when $t=2$ is $-24\text{ft/s}$ .

Step-by-step solution

Step 1. Write the given information.

The velocity is $40\text{ft/s}$, height is $y=40t-16t^2$.

Write the formula used.

The instantaneous rate of change of a function, and the instantaneous velocity of an object at $x=a$ all required us to compute the following limit.$f\text{'}\left(a\right)=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$

Step 3. Find the instantaneous velocity.

The instantaneous velocity of the ball when$t=2$is$y\text{'}\left(2\right)$

We have,

$\begin{array}{c}y\text{'}\left(2\right)=\underset{t\to 2}{\lim }\frac{y\left(t\right)-y\left(2\right)}{t-2}\\ =\underset{t\to 2}{\lim }\frac{(40t-16t^2)-\left\{40\right(2)-16(4\left)\right\}}{t-2}\\ =\underset{t\to 2}{\lim }\frac{(40t-16t^2)-(80-64)}{t-2}\\ =\underset{t\to 2}{\lim }\frac{(40t-16t^2)-16}{t-2}\\ =\underset{t\to 2}{\lim }\frac{-16t^2+40t-16}{t-2}\\ =\underset{t\to 2}{\lim }\frac{(8-16t)(t-2)}{t-2}\\ =\underset{t\to 2}{\lim }(8-16t)\\ =8-16\left(2\right)\\ =8-32\\ =-24\end{array}$

The instantaneous velocity when $t=2$ is$-24\text{ft/s}$ .