Question

topics: limit definition of derivatives, differentiability
note. you are not allowed to use any derivative formulas for these questions. any derivatives you may need to calculate need to be done so using the limit definition.

1. san francisco 49ers quarterback brock purdy is practicing throwing the football in training camp. the quarterback coach for the time has found that the distance the ball travels which in the air until it hits the ground, measured in seconds, is,

d(t) = \frac{t^{3}+25t}{3} metres, 0 \leq t \leq 4
(a) how far has the ball gone when it hits the ground?
(b) calculate the average velocity of the ball between t = 1 and t = 4 seconds.
(c) calculate the velocity function v(t) and the acceleration function a(t) using limits.
(d) determine the instantaneous velocity and acceleration of the ball at t = 3 seconds.

Explanation:

Step1: Find distance when ball hits ground

We are given the time - interval $0\leq t\leq4$. When the ball hits the ground, we assume $t = 4$. Substitute $t = 4$ into the distance function $d(t)=\frac{t^{3}+25t}{3}$.
\[d(4)=\frac{4^{3}+25\times4}{3}=\frac{64 + 100}{3}=\frac{164}{3}\text{ metres}\]

Step2: Calculate average velocity

The average - velocity formula is $v_{avg}=\frac{d(t_2)-d(t_1)}{t_2 - t_1}$. Here, $t_1 = 1$ and $t_2 = 4$. First, find $d(1)$ and $d(4)$.
\[d(1)=\frac{1^{3}+25\times1}{3}=\frac{1 + 25}{3}=\frac{26}{3}\]
\[d(4)=\frac{164}{3}\]
\[v_{avg}=\frac{d(4)-d(1)}{4 - 1}=\frac{\frac{164}{3}-\frac{26}{3}}{3}=\frac{\frac{164 - 26}{3}}{3}=\frac{\frac{138}{3}}{3}=\frac{46}{3}\text{ m/s}\]

Step3: Calculate velocity function using limits

The velocity function $v(t)$ is the derivative of the distance function $d(t)$ using the limit definition $v(t)=\lim_{h
ightarrow0}\frac{d(t + h)-d(t)}{h}$.
\[d(t)=\frac{t^{3}+25t}{3}\]
\[d(t + h)=\frac{(t + h)^{3}+25(t + h)}{3}=\frac{t^{3}+3t^{2}h + 3th^{2}+h^{3}+25t+25h}{3}\]
\[v(t)=\lim_{h
ightarrow0}\frac{\frac{t^{3}+3t^{2}h + 3th^{2}+h^{3}+25t+25h}{3}-\frac{t^{3}+25t}{3}}{h}\]
\[=\lim_{h
ightarrow0}\frac{\frac{t^{3}+3t^{2}h + 3th^{2}+h^{3}+25t+25h-(t^{3}+25t)}{3}}{h}\]
\[=\lim_{h
ightarrow0}\frac{\frac{3t^{2}h + 3th^{2}+h^{3}+25h}{3}}{h}\]
\[=\lim_{h
ightarrow0}\frac{3t^{2}h + 3th^{2}+h^{3}+25h}{3h}\]
\[=\lim_{h
ightarrow0}\frac{h(3t^{2}+3th + h^{2}+25)}{3h}\]
\[=\lim_{h
ightarrow0}\frac{3t^{2}+3th + h^{2}+25}{3}=t^{2}+\frac{25}{3}\]
The acceleration function $a(t)$ is the derivative of the velocity function $v(t)$ using the limit definition $a(t)=\lim_{h
ightarrow0}\frac{v(t + h)-v(t)}{h}$.
\[v(t)=t^{2}+\frac{25}{3}\]
\[v(t + h)=(t + h)^{2}+\frac{25}{3}=t^{2}+2th+h^{2}+\frac{25}{3}\]
\[a(t)=\lim_{h
ightarrow0}\frac{(t^{2}+2th+h^{2}+\frac{25}{3})-(t^{2}+\frac{25}{3})}{h}\]
\[=\lim_{h
ightarrow0}\frac{2th+h^{2}}{h}=\lim_{h
ightarrow0}(2t + h)=2t\]

Step4: Calculate instantaneous velocity and acceleration at $t = 3$

Substitute $t = 3$ into the velocity and acceleration functions.
For velocity: $v(3)=3^{2}+\frac{25}{3}=9+\frac{25}{3}=\frac{27 + 25}{3}=\frac{52}{3}\text{ m/s}$
For acceleration: $a(3)=2\times3 = 6\text{ m/s}^2$

Answer:

(a) $\frac{164}{3}$ metres
(b) $\frac{46}{3}$ m/s
(c) $v(t)=t^{2}+\frac{25}{3}$, $a(t)=2t$
(d) Instantaneous velocity: $\frac{52}{3}$ m/s, Instantaneous acceleration: $6$ m/s²