Question

suppose that a ball is rolling down a ramp. the distance traveled by the ball is given by the function s(t)=9t², where t is the time, in seconds, after the ball is released, and s(t) is measured in feet. find the balls average velocity in each of the following time intervals.
a. t₁ = 4 to t₂ = 5
δs/δt = 81 ft/sec
b. t₁ = 4 to t₂ = 4.5
δs/δt = ft/sec

Explanation:

Step1: Recall average - velocity formula

The average velocity $\frac{\Delta s}{\Delta t}=\frac{s(t_2)-s(t_1)}{t_2 - t_1}$, where $s(t)=9t^{2}$.

Step2: Calculate $s(t_1)$ and $s(t_2)$ for $t_1 = 4$ and $t_2=4.5$

First, find $s(t_1)$ when $t_1 = 4$. Substitute $t_1$ into $s(t)$: $s(4)=9\times4^{2}=9\times16 = 144$. Then, find $s(t_2)$ when $t_2 = 4.5$. Substitute $t_2$ into $s(t)$: $s(4.5)=9\times(4.5)^{2}=9\times20.25 = 182.25$.

Step3: Calculate average velocity

$\frac{\Delta s}{\Delta t}=\frac{s(4.5)-s(4)}{4.5 - 4}=\frac{182.25 - 144}{0.5}=\frac{38.25}{0.5}=76.5$.

Answer:

$76.5$ ft/sec