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

kloseyq 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 \(\displaystyle\alpha={1.50}\ \frac{{m}}{{s}^{{{2}}}}\) and \(\displaystyle\beta={0.0500}\ \frac{{m}}{{s}^{{{3}}}}\)
Calculate the average velocity of the car for each time interval:
a) \(\displaystyle{t}={0}\) to \(\displaystyle{t}={2.00}{s}\)
b) \(\displaystyle{t}={0}\) to \(\displaystyle{t}={4.00}\ {s}\)
c) \(\displaystyle{t}={2.00}{s}\) to \(\displaystyle{t}={4.00}{s}\)

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Expert Answer

chumants6g
Answered 2022-01-13 Author has 4630 answers
Step 1
First we have to calculate the distance x at each t.
\(\displaystyle{x}{\left({0}\right)}=\text{Zero}\ {m}\)
\(\displaystyle{x}{\left({2}\right)}={1.50}{\left({2}\right)}^{{{2}}}-{0.0500}{\left({2}\right)}^{{{3}}}={5.6}{m}\)
\(\displaystyle{x}{\left({4}\right)}={1.50}{\left({4}\right)}^{{{2}}}-{0.0500}{\left({4}\right)}^{{{3}}}={20.8}{m}\)
The average velocity is the displacement divided by the time interval at which this displacement happened.
\(\displaystyle{v}_{{{x},\ {a}{v}{g}}}={\frac{{\Delta{x}}}{{\Delta{t}}}}\)
\(\displaystyle\Delta{x}={x}_{{{f}}}-{x}_{{{i}}}\)
\(\displaystyle\Delta{t}={t}_{{{f}}}-{t}_{{{i}}}\)
Step 2
a) \(\displaystyle{v}_{{{x},\ {a}{v}{g}}}={\frac{{{x}_{{{f}}}-{x}_{{{i}}}}}{{{t}_{{{f}}}-{t}_{{{i}}}}}}\)
\(\displaystyle{x}{\left({0}\right)}=\text{Zero}\ {m}\)
\(\displaystyle{x}{\left({2}\right)}={1.50}{\left({2}\right)}^{{{2}}}-{0.0500}{\left({2}\right)}^{{{3}}}={5.6}{m}\)
\(\displaystyle{v}_{{{x},\ {a}{v}{g}}}={\frac{{{5.6}-{0}}}{{{2}-{0}}}}\)
\(\displaystyle={2.8}{m}{s}\)
Step 3
b) \(\displaystyle{x}{\left({0}\right)}=\text{Zero}\ {m}\)
\(\displaystyle{x}{\left({4}\right)}={1.50}{\left({4}\right)}^{{{2}}}-{0.0500}{\left({4}\right)}^{{{3}}}={20.8}{m}\)
\(\displaystyle{v}_{{{x},\ {a}{v}{g}}}={\frac{{{20.8}-{0}}}{{{4}-{0}}}}\)
\(\displaystyle={5.2}{m}{s}\)
Step 4
c) \(\displaystyle{x}{\left({2}\right)}={1.50}{\left({2}\right)}^{{{2}}}-{0.0500}{\left({2}\right)}^{{{3}}}={5.6}{m}\)
\(\displaystyle{x}{\left({4}\right)}={1.50}{\left({4}\right)}^{{{2}}}-{0.0500}{\left({4}\right)}^{{{3}}}={20.8}{m}\)
\(\displaystyle{v}_{{{x},\ {a}{v}{g}}}={\frac{{{20.8}-{5.6}}}{{{4}-{2}}}}\)
\(\displaystyle={7.6}{m}{s}\)
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Mary Herrera
Answered 2022-01-14 Author has 4444 answers
Step 1
a) The equation for car's distance x from a stop sign is given as a function of time is,
1) \(\displaystyle{x}{\left({t}\right)}=\alpha{t}^{{{2}}}-\beta{t}^{{{3}}}\)
x is the dicplacement of the car
t is the instantaneous time
Let \(\displaystyle{t}_{{{1}}}\) be the initial time for which the initial position of the car is \(\displaystyle{x}_{{{1}}}\)
Substitute 0 for \(\displaystyle{t}_{{{1}}},\ {1.50}\frac{{m}}{{s}^{{{2}}}}\) for \(\displaystyle\alpha\), and \(\displaystyle{0.0500}{m}{s}^{{{3}}}\) for \(\displaystyle\beta\) in equation (1) to find the initial position \(\displaystyle{x}_{{{1}}}\)
\(\displaystyle{x}_{{{1}}}={\left({150}{m}{s}^{{{2}}}{\left({0}\right)}^{{{2}}}\right)}-{\left({0.050}{m}{s}^{{{3}}}{\left({0}\right)}^{{{3}}}\right)}\)
\(\displaystyle={0}\)
Let \(\displaystyle{t}_{{{2}}}\) be the final time for which the final position of the car is \(\displaystyle{x}_{{{2}}}\)
Substitute 2.00 s for \(\displaystyle{t}_{{{2}}},\ {1.50}{m}{s}^{{{2}}}\) for \(\displaystyle\alpha,\ {0.0500}{m}{s}^{{{3}}}\) for \(\displaystyle\beta\) in equation (1) to find the final position \(\displaystyle{x}_{{{2}}}\)
\(\displaystyle{x}_{{{2}}}={\left({1.50}\ \frac{{m}}{{s}^{{{2}}}}{\left({2.00}{s}\right)}^{{{2}}}\right)}-{\left({0.050}\ \frac{{m}}{{s}^{{{3}}}}{\left({2.00}{s}\right)}^{{{3}}}\right)}\)
\(\displaystyle={5.60}{m}\)
Formula to calculate the displacement is,
\(\displaystyle\Delta{x}={x}_{{{2}}}-{x}_{{{1}}}\)
\(\displaystyle\Delta{x}\) is the total displacement between the position \(\displaystyle{x}_{{{1}}}\) and \(\displaystyle{x}_{{{2}}}\)
Substitute \(\displaystyle{5.60}{m}\) for \(\displaystyle{x}_{{{2}}}\), and 0m for \(\displaystyle{x}_{{{1}}}\) to find \(\displaystyle\Delta{x}\)
\(\displaystyle\Delta{x}={\left({5.60}{m}-{0}\right)}\)
\(\displaystyle={5.60}{m}\)
Thus, the displacement of the car between the time interval \(\displaystyle{t}={0}\) to \(\displaystyle{t}={2.00}{s}\) is 5.60 m
Formula to calculate the total time is,
\(\displaystyle\Delta{t}={t}_{{{2}}}-{t}_{{{1}}}\)
\(\displaystyle\Delta\) is the total time displacement between the position \(\displaystyle{x}_{{{1}}}\) and \(\displaystyle{x}_{{{2}}}\)
Substitute 0s for \(\displaystyle{t}_{{{1}}}\), and 2.00s for \(\displaystyle{t}_{{{2}}}\) to find \(\displaystyle\Delta\)
\(\displaystyle\Delta{t}={2.00}{s}-{0}\)
\(\displaystyle={2.00}{s}\)
Thus, the time taken by the car to move from initial displacement \(\displaystyle{x}_{{{1}}}\) to final displacement \(\displaystyle{x}_{{{2}}}\) is 2.00s
Conclusion:
Formula to calculate the average velocity is,
\(\displaystyle{v}_{{{a}{v}-{x}}}={\frac{{\Delta{x}}}{{\Delta{t}}}}\)
\(\displaystyle{v}_{{{a}{v}-{x}}}\) is the average velocity of the car.
Substitute 5.60m for \(\displaystyle\Delta{x},\ {2.00}{s}\) for \(\displaystyle\Delta\) to find \(\displaystyle{v}_{{{a}{v}-{x}}}\)
\(\displaystyle{v}_{{{a}{v}-{x}}}={\frac{{{5.60}{m}}}{{{2.00}{s}}}}\)
\(\displaystyle={2.80}\frac{{m}}{{s}}\)
Therefore, the average velocity of the ca interval \(\displaystyle{t}={0}{s}\) to \(\displaystyle{t}={2.00}{s}\) is \(\displaystyle{2.80}\frac{{m}}{{s}}\)
Step 2
b) Substitute 0s for \(\displaystyle{t}_{{{1}}}\), and 4.00s for \(\displaystyle{t}_{{{2}}}\) to find \(\displaystyle\Delta\)
\(\displaystyle\Delta{t}={4.00}{s}-{0}\)
\(\displaystyle={4.00}{s}\)
Thus, the time taken by the car to move from initial displacement \(\displaystyle{x}_{{{1}}}\) to final displacement \(\displaystyle{x}_{{{2}}}\) is 4.00s
Formula to calculate the average velocity is,
\(\displaystyle{v}_{{{a}{v}-{x}}}={\frac{{\Delta{x}}}{{\Delta{t}}}}\)
Substitute 20.8m for \(\displaystyle\Delta{x},\ {4.00}{s}\) for \(\displaystyle\Delta\) to find \(\displaystyle{v}_{{{a}{v}-{x}}}\)
\(\displaystyle{v}_{{{a}{v}-{x}}}={\frac{{{20.8}{m}}}{{{4.00}{s}}}}\)
\(\displaystyle={5.20}\frac{{m}}{{s}}\)
Therefore, the average velocity of the car interval \(\displaystyle{t}={0}{s}\) to \(\displaystyle{t}={4.00}{s}\) is \(\displaystyle{5.20}\frac{{m}}{{s}}\)
Step 3
c) From equation (1) of (a)
1) \(\displaystyle{x}{\left({t}\right)}=\alpha{t}^{{{2}}}-\beta{t}^{{{3}}}\)
\(\displaystyle{x}_{{{1}}}={\left({1.50}\frac{{m}}{{s}^{{{2}}}}{\left({2.00}{s}\right)}^{{{2}}}\right)}-{\left({0.050}\frac{{m}}{{s}^{{{3}}}}{\left({2.00}{s}\right)}^{{{3}}}\right)}\)
\(\displaystyle={5.60}{m}\)
\(\displaystyle{x}_{{{2}}}={\left({1.50}\frac{{m}}{{s}^{{{2}}}}{\left({4.00}{s}\right)}^{{{2}}}\right)}-{\left({0.050}\frac{{m}}{{s}^{{{3}}}}{\left({4.00}{s}\right)}^{{{3}}}\right)}\)
\(\displaystyle={20.8}{m}\)
\(\displaystyle\Delta{x}={\left({20.8}{m}-{5.60}{m}\right)}\)
\(\displaystyle={15.2}{m}\)
\(\displaystyle\Delta{t}={4.00}{s}-{2.00}{s}\)
\(\displaystyle={2.00}{s}\)
\(\displaystyle{v}_{{{a}{v}-{x}}}={\frac{{{15.2}{m}}}{{{2.00}{s}}}}\)
\(\displaystyle={7.60}{m}{s}\)
Therefore, the average velocity of the car for interval \(\displaystyle{t}={2.00}{s}\) to \(\displaystyle{t}={4.00}{s}\) is \(\displaystyle{7.60}\frac{{m}}{{s}}\)
0
nick1337
Answered 2022-01-14 Author has 10160 answers

Step 1
a) The position of the car as a function of time t is given by
\(x(t)=\alpha t^{2}-\beta t^{3}\)
where
\(\alpha=1.50m/s^{2} \\\beta=0.05m/s^{3}\)
The average velocity is given by the ratio between the displacement and the time taken:
\(v=\frac{\Delta x}{\Delta t}\)
The position at t = 0 is:
\(x(0)=\alpha\times0^{2}-\beta\times0^{3}=0\)
The position at t = 2.00 s is:
\(x(2)=\alpha\times2^{2}-\beta\times2^{3}=5.6m\)
So the displacement is
\(\Delta x=x(2)-x(0)=5.6-0=5.6m\)
The time interval is
\(\Delta t=2.0s-0s=2.0s\)
And so, the average velocity in this interval is
\(v=\frac{5.6m}{2.0s}=2.8m/s\)
b) The position at t = 0 is:
\(x(0)=\alpha\times0^{2}-\beta\times0^{3}=0\)
While the position at t = 4.00 s is:
\(x(4)=\alpha\times4^{2}-\beta\times4^{3}=20.8m\)
So the displacement is
\(\Delta x=x(4)-x(0)=20.8-0=20.8m\)
The time interval is
\(\Delta t=4.00-0=4.0s\)
So the average velocity here is
\(v=\frac{20.8}{4.0}=5.2m/s\)
c) The position at t = 2 s is:
\(x(2)=\alpha\times2^{2}-\beta\times2^{3}=5.6m\)
While the position at t = 4 s is:

\(x(4)=\alpha\times4^{2}-\beta\times4^{3}=20.8m\)
So the displacement is
\(\Delta x=20.8-5.6=15.2m\)
While the time interval is
\(\Delta t=4.0-2.0=2.0s\)
So the average velocity is
\(v=\frac{15.2}{2.0}=7.6m/s\)

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