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 \(\displaystyle{7.60}\frac{{m}}{{s}}\)