Part a.
i. -20.6(m/s).
ii. -19.16(m/s).
iii. -19.016(m/s).
Part b.
-19m/s.
Let's refer to the average velocity formula:
[tex]A_v=\frac{P_2-P_1}{t_2-t_1}[/tex]; where [tex]P_2[/tex] and [tex]P_1[/tex] are the final position, and initial position, respectively. Also, [tex]t_2[/tex] and [tex]t_1[/tex] are the final time, and initial time, respectively.
[tex]P_1=45(2)-16(2)^2=26\\ \\P_2=45(2.1)-16(2.1)^2=23.94[/tex]
[tex]A_v=\frac{P_2-P_1}{t_2-t_1}=\frac{23.94m-26m}{2.1s-2s} =-20.6m/s.[/tex]
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[tex]P_1=45(2)-16(2)^2=26\\ \\P_2=45(2.01)-16(2.01)^2=25.8084[/tex]
[tex]A_v=\frac{P_2-P_1}{t_2-t_1}=\frac{25.8084m-26m}{2.01s-2s} =-19.16m/s.[/tex]
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[tex]P_1=45(2)-16(2)^2=26\\ \\P_2=45(2.001)-16(2.001)^2= 25.980984[/tex]
[tex]A_v=\frac{P_2-P_1}{t_2-t_1}=\frac{25.980984m-26m}{2.001s-2s} =-19.016m/s.[/tex]
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Formula of instantaneous velocity:
[tex]V_I=\frac{dx}{dt} .[/tex]
The instantaneous velocity indicates that we need to find the derivative with respect to time (t) of the position equation, and evaluate it for the indicated time. Let's do that:
[tex]y'=(1)45t^{1-1} -(2)16t^{2-1} \\ \\y'=45-32t[/tex]
Now, we evaluate for the indicated time (t= 2).
[tex]y'=45-32(2)=-19m/s.[/tex]
Hence, the instantaneous velocity for t= 2 is -19m/s.
Answer:
a) i) -20.6 ft/s
ii) -19.16 ft/s
iii) -19.016 ft/s
b) -19 ft/s
Step-by-step explanation:
Given information:
[tex]\boxed{\begin{minipage}{4 cm}\underline{Average velocity formula}\\\\ $\overline{v}=\dfrac{s(t_2)-s(t_1)}{t_2-t_1}$\\\end{minipage}}[/tex]
where:
Question (i)
Given interval: [2, 2.1]
[tex]\implies t_1=2[/tex]
[tex]\implies t_2=2.1[/tex]
[tex]\implies s(t_1)=45(2)-16(2)^2=26[/tex]
[tex]\implies s(t_2)=45(2.1)-16(2.1)^2=23.94[/tex]
Substitute the values into the formula and solve for v:
[tex]\begin{aligned}\overline{v} & =\dfrac{s(t_2)-s(t_1)}{t_2-t_1}\\\\\implies \overline{v} & =\dfrac{s(2.1)-s(2)}{2.1-2}\\\\& =\dfrac{23.94-26}{2.1-2}\\\\& =\dfrac{-2.06}{0.1}\\\\& =-20.6\:\: \sf ft/s\end{aligned}[/tex]
Question (ii)
Given interval: [2, 2.01]
[tex]\implies t_1=2[/tex]
[tex]\implies t_2=2.01[/tex]
[tex]\implies s(t_1)=45(2)-16(2)^2=26[/tex]
[tex]\implies s(t_2)=45(2.01)-16(2.01)^2=25.8084[/tex]
Substitute the values into the formula and solve for v:
[tex]\begin{aligned}\overline{v} & =\dfrac{s(t_2)-s(t_1)}{t_2-t_1}\\\\\implies \overline{v} & =\dfrac{s(2.01)-s(2)}{2.01-2}\\\\& =\dfrac{25.8084-26}{2.01-2}\\\\& =\dfrac{-0.1916}{0.01}\\\\& =-19.16\:\: \sf ft/s\end{aligned}[/tex]
Question (iii)
Given interval: [2, 2.001]
[tex]\implies t_1=2[/tex]
[tex]\implies t_2=2.001[/tex]
[tex]\implies s(t_1)=45(2)-16(2)^2=26[/tex]
[tex]\implies s(t_2)=45(2.001)-16(2.001)^2=25.980984[/tex]
Substitute the values into the formula and solve for v:
[tex]\begin{aligned}\overline{v} & =\dfrac{s(t_2)-s(t_1)}{t_2-t_1}\\\\\implies \overline{v} & =\dfrac{s(2.001)-s(2)}{2.001-2}\\\\& =\dfrac{25.980984-26}{2.001-2}\\\\& =\dfrac{-0.019016}{0.001}\\\\& =-19.016\:\: \sf ft/s\end{aligned}[/tex]
[tex]\boxed{\begin{minipage}{5cm}\underline{Instantaneous velocity formula}\\\\$v_i=\lim_{\triangle t \to 0} \dfrac{\text{d}s}{\text{d}t}$\end{minipage}}[/tex]
To find the equation for instantaneous velocity, differentiate the function for height:
[tex]\begin{aligned}s(t) & =45t-16t^2\\\implies \dfrac{\text{d}s}{\text{d}t} & = 1 \cdot 45t^{1-1} - 2 \cdot 16t^{2-1}\\v_i& = 45-32t\end{aligned}[/tex]
To find the instantaneous velocity when t = 2, substitute t = 2 into the found instantaneous velocity equation:
[tex]\begin{aligned}v_i & =45-32(t)\\t=2\implies v_i & = 45-32(2)\\& = 45-64\\& = -19\:\: \sf ft/s\end{aligned}[/tex]
[tex]\boxed{\begin{minipage}{4.8 cm}\underline{Differentiating $ax^n$}\\\\If $y=ax^n$, then $\dfrac{\text{d}y}{\text{d}x}=nax^{n-1}$\\\end{minipage}}[/tex]
