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

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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GaceCoect5v
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) $$\displaystyle{x}{\left({2}\right)}={1.50}{\left({2}\right)}^{{{2}}}-{0.0500}{\left({2}\right)}^{{{3}}}={5.6}{m}$$
$$\displaystyle{x}{\left({4}\right)}={1.50}{\left({4}\right)}^{{{2}}}-{0.0500}{\left({4}\right)}^{{{3}}}={20.8}{m}$$
$$\displaystyle{v}_{{{x},{a}{v}{g}}}={\frac{{{20.8}-{5.6}}}{{{4}-{2}}}}$$
$$\displaystyle={7.6}\frac{{m}}{{s}}$$
###### Not exactly what you’re looking for?
movingsupplyw1

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
$$\displaystyle{v}={\frac{{{15.2}}}{{{2.0}}}}={7.6}\frac{{m}}{{s}}$$

nick1337

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