Question

Pls help w this pretty urgent

Answer 1

Answer:

$\begin{aligned} \textsf{a)} \quad & \textsf{Position function}: \quad s(t) = -16t^2+1362 \\ & \textsf{Velocity function}: \quad v(t) = -32t\end{aligned}$

$\textsf{b)} \quad \overline{v}=-48\:\: \sf ft/s$

$\begin{aligned}\textsf{c)} \quad v(1) & = -32\:\: \sf ft/s\\ v(2) & = -64\:\: \sf ft/s\end{aligned}$

$\textsf{d)} \quad t = 9.226\:\: \sf s \:\: (3\: d.p.)$

$\textsf{e)} \quad v = -292.242\:\: \sf ft/s \:\: (3\: d.p.)$

Step-by-step explanation:

Given:

$s(t)=-16t^2+v_0t+s_0$

Part (a)

If v₀ is the initial velocity, and s₀ is the initial position:

$\implies v_0=0$

$\implies s_0=1362$

Therefore:

$\begin{aligned}\textsf{Position function}: \quad s(t) & =-16t^2+0+1362\\ \implies s(t) & = -16t^2+1362 \end{aligned}$

$\begin{aligned}\textsf{Velocity function}: \quad v(t)&=s\:'(t)\\ \implies v(t) &=-32t\end{aligned}$

Part (b)

Average velocity is the change in displacement divided by the change in time:

$\overline{v}=\dfrac{s(t_2)-s(t_1)}{t_2-t_1}$

Given interval: 1 ≤ t ≤ 2

$\implies t_1=1$

$\implies t_2=2$

Find the values of s(t) at t = 1 and t = 2:

$\implies s(1)=-16(1)^2+1362=1346$

$\implies s(2)=-16(2)^2+1362=1298$

Substitute the values into the formula:

$\overline{v}=\dfrac{s(2)-s(1)}{2-1}=\dfrac{1298-1346}{2-1}=-48\:\: \sf ft/s$

Part (c)

To find the velocity at t = 1 and t = 2, substitute these values into the velocity function found in part (a):

$\implies v(1)=-32(1)=-32\:\: \sf ft/s$

$\implies v(2)=-32(2)=-64\:\: \sf ft/s$

Part (d)

The time required for the coin to reach ground level is when s(t) = 0:

$\begin{aligned}s(t) & = 0\\-16t^2+1362 & = 0\\16t^2 & = 1362\\t^2 & = 85.125\\t & = \sqrt{85.125}\\t & = 9.226\: \sf s\:\:(3\:d.p.)\end{aligned}$

Part (e)

To find the velocity of the coin at impact with the ground, substitute the found value of t from part (d) into the equation for velocity found in part (a):

$\begin{aligned}v(t) & = -32t\\\implies v(\sqrt{85.125}) & = -32(\sqrt{85.125})\\ v & =-295.242\:\: \sf ft/s\:\:(3\:d.p.)\end{aligned}$

Note: I have used the non-rounded value of t from part (d) for accuracy.