A Honda Civic travels in a straight line along a road. The car’s dista

amolent3u 2022-01-12 Answered
A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by the equation \(\displaystyle{x}{\left({t}\right)}=\alpha{t}^{{{2}}}-\beta{t}^{{{3}}}\), where \(\displaystyle\alpha={1.50}\frac{{m}}{{s}^{{{2}}}}\ {\quad\text{and}\quad}\ \beta={0.0500}\frac{{m}}{{s}^{{{3}}}}\). Calculate the average velocity of the car for each time interval: (a) \(\displaystyle{t}={0}\ \to\ {t}={2.00}{s}\); (b) \(\displaystyle{t}={0}\ \to\ {t}={4.00}{s}\); (c) \(\displaystyle{t}={2.00}{s}\ \to\ {t}={4.00}{s}\).

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Expert Answer

GaceCoect5v
Answered 2022-01-13 Author has 3528 answers
Step 1
First we have to calculate the distance x at each t.
\(\displaystyle{x}{\left({0}\right)}={Z}{e}{r}{o}\ {m}\)
\(\displaystyle{x}{\left({2}\right)}={1.50}{\left({2}\right)}^{{{2}}}-{0.0500}{\left({2}\right)}^{{{3}}}={5.6}{m}\)
\(\displaystyle{x}{\left({4}\right)}={1.50}{\left({4}\right)}^{{{2}}}-{0.0500}{\left({4}\right)}^{{{3}}}={20.8}{m}\)
The average velocity is the displacement divided by the time interval at which this displacement happened.
\(\displaystyle{v}_{{{x},{a}{v}{g}}}={\frac{{\triangle{x}}}{{\triangle{t}}}}\)
\(\displaystyle\triangle{x}={x}_{{{f}}}-{x}_{{{i}}}\)
\(\displaystyle\triangle{t}={t}_{{{f}}}-{t}_{{{i}}}\)
Step 2
(a) \(\displaystyle{v}_{{{x},{a}{v}{g}}}={\frac{{{x}_{{{f}}}-{x}_{{{i}}}}}{{{t}_{{{f}}}-{t}_{{{i}}}}}}\)
\(\displaystyle{x}{\left({0}\right)}={Z}{e}{r}{o}\ {m}\)
\(\displaystyle{x}{\left({2}\right)}={1.50}{\left({2}\right)}^{{{2}}}-{0.0500}{\left({2}\right)}^{{{3}}}={5.6}{m}\)
\(\displaystyle{v}_{{{x},{a}{v}{g}}}={\frac{{{5.6}-{0}}}{{{2}-{0}}}}\)
\(\displaystyle={2.8}\frac{{m}}{{s}}\)
Step 3
(b) \(\displaystyle{x}{\left({0}\right)}={Z}{e}{r}{o}\ {m}\)
\(\displaystyle{x}{\left({4}\right)}={1.50}{\left({4}\right)}^{{{2}}}-{0.0500}{\left({4}\right)}^{{{3}}}={20.8}{m}\)
\(\displaystyle{v}_{{{x},{a}{v}{g}}}={\frac{{{20.8}-{0}}}{{{4}-{0}}}}\)
\(\displaystyle={5.2}\frac{{m}}{{s}}\)
Step 4
(c) \(\displaystyle{x}{\left({2}\right)}={1.50}{\left({2}\right)}^{{{2}}}-{0.0500}{\left({2}\right)}^{{{3}}}={5.6}{m}\)
\(\displaystyle{x}{\left({4}\right)}={1.50}{\left({4}\right)}^{{{2}}}-{0.0500}{\left({4}\right)}^{{{3}}}={20.8}{m}\)
\(\displaystyle{v}_{{{x},{a}{v}{g}}}={\frac{{{20.8}-{5.6}}}{{{4}-{2}}}}\)
\(\displaystyle={7.6}\frac{{m}}{{s}}\)
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movingsupplyw1
Answered 2022-01-14 Author has 6048 answers

Explanation:
a) The position of the car as a function of time t is given by
\(\displaystyle{x}{\left({t}\right)}=\alpha{t}^{{{2}}}-\beta{t}^{{{3}}}\)
where \(\displaystyle\alpha={1.50}\frac{{m}}{{s}^{{{2}}}}\)
\(\displaystyle\beta={0.05}\frac{{m}}{{s}^{{{3}}}}\)
The average velocity is given by the ratio between the displacement and the time taken:
\(\displaystyle{v}={\frac{{\triangle{x}}}{{\triangle{t}}}}\)
The position at \(\displaystyle{t}={0}\) is:
\(\displaystyle{x}{\left({0}\right)}=\alpha\cdot{0}^{{{2}}}-\beta\cdot{0}^{{{3}}}={0}\)
The position at \(\displaystyle{t}={2.00}\) s is:
\(\displaystyle{x}{\left({2}\right)}=\alpha\cdot{2}^{{{2}}}-\beta\cdot{2}^{{{3}}}={5.6}{m}\)
So the displacement is
\(\displaystyle\triangle{x}={x}{\left({2}\right)}-{x}{\left({0}\right)}={5.6}-{0}={5.6}{m}\)
The time interval is
\(\displaystyle\triangle{t}={2.0}{s}-{0}{s}={2.0}{s}\)
And so, the average velocity in this interval is
\(\displaystyle{v}={\frac{{{5.6}{m}}}{{{2.0}{s}}}}={2.8}\frac{{m}}{{s}}\)
b) The position at \(\displaystyle{t}={0}\) is:
\(\displaystyle{x}{\left({0}\right)}=\alpha\cdot{0}^{{{2}}}-\beta\cdot{0}^{{{3}}}={0}\)
While the position at \(\displaystyle{t}={4.00}\) s is:
\(\displaystyle{x}{\left({4}\right)}=\alpha\cdot{4}^{{{2}}}-\beta\cdot{4}^{{{3}}}={20.8}{m}\)
So the displacement is
\(\displaystyle\triangle{x}={x}{\left({4}\right)}-{x}{\left({0}\right)}={20.8}-{0}={20.8}{m}\)
The time interval is
\(\displaystyle\triangle{t}={4.0}-{0}={4.0}{s}\)
So the average velocity here is
\(\displaystyle{v}={\frac{{{20.8}}}{{{4.0}}}}={5.2}\frac{{m}}{{s}}\)
c) The position at \(\displaystyle{t}={2}\) s is:
\(\displaystyle{x}{\left({2}\right)}=\alpha\cdot{2}^{{{2}}}-\beta\cdot{2}^{{{3}}}={5.6}{m}\)
While the position at \(\displaystyle{t}={4}{s}\) is:
\(\displaystyle{x}{\left({4}\right)}=\alpha\cdot{4}^{{{2}}}-\beta\cdot{4}^{{{3}}}={20.8}{m}\)
So the displacement is
\(\displaystyle\triangle{x}={20.8}-{5.6}={15.2}{m}\)
While the time interval is
\(\displaystyle\triangle{t}={4.0}-{2.0}={2.0}{s}\)
So the average velocity is
\(\displaystyle{v}={\frac{{{15.2}}}{{{2.0}}}}={7.6}\frac{{m}}{{s}}\)

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nick1337
Answered 2022-01-14 Author has 10160 answers

\(x(t)=\alpha t^{2}-\beta t^{3} \\\alpha=1.64 m/s^{2} \\\beta=0.0521 m/s^{2} \\t_{0}=2s \\t_{1}=4S \\V_{avg}=\frac{\triangle x}{\triangle t}=\frac{[x(t_{1})-x(t_{0})]}{(t_{1}-t_{0})} \\x(t_{1})=x(4)=1.64 \times 4^{2}-0.0521 \times 4^{3} \\=22.90m \\X(t_{0})=x(2)=1.64 \times 2^{2}-0.0521 \times 2^{3} \\=6.143m \\V_{avg}=\frac{22.90-6.143}{4-2}=8.378 m/s\)

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