For an uniformly accelerated motion, the following relationship can be used:
[tex]2aS = v_f^2 -v_i^2[/tex]
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:
[tex]S=4 km=4000 m[/tex]
To convert the speed from km/h to m/s, let's keep in mind that
[tex]1 km=1000 m[/tex]
[tex]1h=3600 s[/tex]
so
[tex]1 \frac{km}{h}= \frac{1000 m}{3600 s}= \frac{1}{3.6} [/tex]
is the conversion factor.
So the speeds are:
[tex]v_i = 15 km/h \cdot \frac{1}{3.6}=4.17 m/s [/tex]
[tex]v_f = 75 km/h \cdot \frac{1}{3.6}=20.83 m/s [/tex]
and so, now we can re-arrange the formula to find a, the average acceleration:
[tex]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 [/tex]
