Question

a bicyclist, steadily speeds up from rest to 17.0mph in 5.40s. how far did she travel (in given units)? ft what was her acceleration (in given units)? ft/s²

Explanation:

Step1: Convert the final velocity from mph to ft/s

We know that \(1\space mph=\frac{5280}{3600}\space ft/s = 1.4667\space ft/s\).
So, \(v = 17.0\space mph\times1.4667\space ft/s/mph=24.934\space ft/s\)

Step2: Find the acceleration

Using the formula \(a=\frac{v - u}{t}\), where \(u = 0\space ft/s\) (starts from rest), \(v=24.934\space ft/s\), and \(t = 5.40\space s\)
\(a=\frac{24.934 - 0}{5.40}=4.62\space ft/s^{2}\)

Step3: Find the distance traveled

Using the formula \(s=ut+\frac{1}{2}at^{2}\), since \(u = 0\space ft/s\), the formula simplifies to \(s=\frac{1}{2}at^{2}\)
Substitute \(a = 4.62\space ft/s^{2}\) and \(t = 5.40\space s\)
\(s=\frac{1}{2}\times4.62\times(5.40)^{2}\)
\(s = 2.31\times29.16=67.36\space ft\)

Answer:

Distance: \(67.4\space ft\)
Acceleration: \(4.62\space ft/s^{2}\)