Question

a bicyclist in the Tour De France crests a mountain pass as he moves at 15km/h. At the bottom 4km/h farther, his speed is 75km/h. What is his average acceleration(m/s^2) while riding down the mountain?,

Answer 1

For an uniformly accelerated motion, the following relationship can be used:
$2aS = v_f^2 -v_i^2$
where a is the acceleration, S the distance covered, vf the final speed and vi the initial speed.

Before using the formula, we need to convert everything into SI units.
The distance covered is:
$S=4 km=4000 m$
To convert the speed from km/h to m/s, let's keep in mind that
$1 km=1000 m$
$1h=3600 s$
so
$1 \frac{km}{h}= \frac{1000 m}{3600 s}= \frac{1}{3.6}$
is the conversion factor.
So the speeds are:
$v_i = 15 km/h \cdot \frac{1}{3.6}=4.17 m/s$
$v_f = 75 km/h \cdot \frac{1}{3.6}=20.83 m/s$

and so, now we can re-arrange the formula to find a, the average acceleration:
$a= \frac{v_f^2-v_i^2}{2S} = \frac{(20.83m/s)^2-(4.17m/s)^2}{2\cdot 4000 m}=0.05 m/s^2$