Problem 2-86 Linear Kinematics - Part 4 - A

A bicyclist, traveling at \(4.0\; km/h\) at the top of a hill coasts downward with constant acceleration, reaching a speed of \(33 \;km/h\) in \(33 \;s.\) What distance, in metres, does the cyclist travel in that time?

Accumulated Solution

\(v_0 = 4 \frac{km}{h} \times \frac{10^2}{km}\times \frac{1\; h}{3600\; s} = 1.11\; m/s \\ v = 33\frac{km}{h} = \frac{1.11}{4} 33 = 9.17 \; m/s\)

Correct.

So a is given by:
\(v = v_0 + at \\ 9.17 = 1.11 + a(33) \\ a = 0.244\; m/s^2\)

The distance traveled is given by:
 

(A)   \(v = v_0 + at\)

(B)   \(x = v_{0^t} + 1/2 at^2\)

(C)   \(v^2 = v_{0^2} + 2ax\)