Show that for a particle moving with velocity $v(t)$, if $v(t).v'(t)=0$ (dot product of $v(t)$ and $v'(t)$ is 0) for all $t$ then the speed $v$ is constant.
I don't really understand this. Isn't it sufficient to say that if $v'(t)$=0, i.e. acceleration is 0, the speed $v$ is constant?
Best Answer
If $v'(t)=0$ for all $t$, then $v(t)$ is constant (implying the speed is constant), but we are not given that $v'(t)=0$ for all $t$. We are only given that $v(t)\cdot v'(t)=0$ for all $t$, which is a weaker condition.
To show the desired result, here is a hint: it suffices to show that the squared speed is constant. The squared speed is $\left\|v(t)\right\|^2 = v(t)\cdot v(t)$. Try and differentiate this with respect to $t$ (use the 'product rule' to help). What do you find the derivative is, and what can you conclude?