# Problem 2-86 Linear Kinematics - Part 3 - B

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.

In fact both methods are correct and you can follow two correct paths to the answer. Let's follow choice (A) first by finding the acceleration.
From the data given, the acceleration can be found from:

(A)   $v = v_0 + at$

(B)   $x = v_{0^t} + 1/2 \;at^2$

(C)   $v^2 = v_{0^2} + 2ax$