Question

a car, initially traveling 28.0ft/s, steadily speeds up to 46.0ft/s in 5.50s. determine all unknowns and answer the following question.
how far did the car travel during this time?

Explanation:

Step1: Find the acceleration

The formula for acceleration \(a=\frac{v - u}{t}\), where \(u = 28.0\mathrm{ft/s}\), \(v=46.0\mathrm{ft/s}\), \(t = 5.50\mathrm{s}\)
\(a=\frac{46.0 - 28.0}{5.50}=\frac{18}{5.50}\approx3.27\mathrm{ft/s^{2}}\)

Step2: Use the displacement formula

The formula for displacement \(s=ut+\frac{1}{2}at^{2}\)
Substitute \(u = 28.0\mathrm{ft/s}\), \(a\approx3.27\mathrm{ft/s^{2}}\), \(t = 5.50\mathrm{s}\)
\(s=(28.0\times5.50)+\frac{1}{2}\times3.27\times(5.50)^{2}\)
First term: \(28.0\times5.50 = 154\)
Second term: \(\frac{1}{2}\times3.27\times30.25= 49.3\)
\(s=154 + 49.3=203.3\mathrm{ft}\)

Another way:

Step1: Use the average - velocity formula

The average velocity \(\overline{v}=\frac{u + v}{2}\), where \(u = 28.0\mathrm{ft/s}\), \(v = 46.0\mathrm{ft/s}\)
\(\overline{v}=\frac{28.0+46.0}{2}=\frac{74}{2}=37\mathrm{ft/s}\)

Step2: Calculate the displacement

The formula \(s=\overline{v}\times t\), with \(\overline{v}=37\mathrm{ft/s}\), \(t = 5.50\mathrm{s}\)
\(s=37\times5.50 = 203.5\mathrm{ft}\) (slight difference due to rounding in the first - method)

Answer:

\(204\mathrm{ft}\) (rounded to three significant figures)