Answer
The magnitude of the rider's acceleration is $~~0.28~m/s^2$
Work Step by Step
We can convert $30~km/h$ to units of $m/s$:
$(30~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 8.33~m/s$
We can find the rider's acceleration:
$v = v_0+at$
$a = \frac{v-v_0}{t}$
$a = \frac{8.33~m/s-0}{30~s}$
$a = 0.28~m/s^2$
The magnitude of the rider's acceleration is $~~0.28~m/s^2$.
