## Engineering Mechanics: Statics & Dynamics (14th Edition)

The acceleration of the particle at s = 15 m is equal to 13.1 $m/s^2$
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: $$a = -2(15) + 0.005(15)^3 = -13.1 \space m/s^2$$