Answer
The acceleration of the particle at s = 15 m is equal to 13.1 $m/s^2$
Work Step by Step
1. Using equation 12-3:
$$a \space ds = v \space dv$$ $$a = v \space \frac{dv}{ds}$$
2. Calculate $dv/ds$:
$$\frac{d(20 - 0.05s^2)}{ds} = -0.1s$$
3. Substitute and find a:
$$a= (20 - 0.05s^2)(-0.1s) = -2s + 0.005s^3$$
At s = 15 m:
$$a = -2(15) + 0.005(15)^3 = -13.1 \space m/s^2$$
