Problem 2-86 Linear Kinematics - Part 5 - 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

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

Correct.

In fact either:

\(x = v_{0^t} + 1/2\; at^2\)
or
\(v^2 = v_{0^2} + 2ax\)
can be used.

\(x = v_{0^t} + 1/2\; at^2 \\ x = 1.11(33) + (1/2)(0.244)(33)^2 = 1.7 \times 10^2\; m\)

Now let us examine the method without having to evaluate the acceleration explicitly.
Here is a statement:

For constant acceleration the displacement \(x\) is given by:

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

(A)   True

(B)   False