# 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 $$\displaystyle{t}={4.00}{s}$$

### Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

### Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it

## 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) $$\displaystyle{x}{\left({2}\right)}={1.50}{\left({2}\right)}^{{{2}}}-{0.0500}{\left({2}\right)}^{{{3}}}={5.6}{m}$$
$$\displaystyle{x}{\left({4}\right)}={1.50}{\left({4}\right)}^{{{2}}}-{0.0500}{\left({4}\right)}^{{{3}}}={20.8}{m}$$
$$\displaystyle{v}_{{{x},\ {a}{v}{g}}}={\frac{{{20.8}-{5.6}}}{{{4}-{2}}}}$$
$$\displaystyle={7.6}{m}{s}$$
###### Not exactly what you’re looking for?
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 $$\displaystyle{7.60}\frac{{m}}{{s}}$$
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
$$v=\frac{15.2}{2.0}=7.6m/s$$

### Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

### Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it
...