Problem 2-86 Linear Kinematics - Part 6 - 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?

diagram of a cyclist on a hill

Accumulated Solution

\(v_0 = 4 \frac{km}{h} \times \frac{10^2}{km}\times \frac{1\; h}{3600\; s} = 1.11\; m/s \\ v = 33\frac{km}{h} = \frac{1.11}{4} 33 = 9.17 \; m/s \\ a = 0.244\; m/s ^2 \\ x = 1.7 \times 10^2\; m\)

Correct.

\(x = v_{av^t}\)

Which is correct:

(A)   \(v_{av} = (1/2)(v - v_0)\)

(B)   \(v_{av} = v + v_0\)

(C)   \(v_{v }= (1/2)(v + v_0)\)