Velocity and Acceleration — Question 6

Question 6

Prove that a particle with nonzero velocity is speeding up exactly when $\vec v\cdot\vec a>0$, and slowing down exactly when $\vec v\cdot\vec a<0$.

Question 6 – Solution

Strategy Differentiate speed through its square to avoid differentiating a square root prematurely.

Derivation Let $v=\|\vec v\|>0$. Then $$v^2=\vec v\cdot\vec v.$$ Differentiating gives $2v\,dv/dt=2\vec v\cdot\vec a$, so $$\boxed{\frac{dv}{dt}=\frac{\vec v\cdot\vec a}{\|\vec v\|}}.$$

Conclusion The denominator is positive. Therefore the sign of the speed derivative is exactly the sign of $\vec v\cdot\vec a$: positive means speeding up, negative means slowing down, and zero means instantaneously constant speed.