I am learning kinematics with vector analysis. I was given the position equation:$\mathbf{r} = 10t\hat{\mathtt{i}} + (20t-5t^2)\hat{\mathtt{j}}$. It asks me the time when the velocity of the particle will be perpendicular to its initial velocity. The teacher taught us how to solve it, but I didn't get the concept. Can someone explain the concept? I remembered him solving this problem by using either dot product or cross product of vector. I expect the explanation will not be too difficult for high school student.
[Physics] When will the velocity of a particle be perpendicular to it’s initial velocity
geometryhomework-and-exerciseskinematicsvectors
Best Answer
Actually @Ocelo7 has already answered it; I am just showing that in a concise way:
Velocity can be found by differentiating the position-vector:$$\dot{\mathbf{r}}= \mathbf{v}(t) = 10\mathtt{\hat{ i}} + (20 - 10t)\mathtt{\hat{j}}$$ .
Assuming the initial time is $t_0$ & the required time is $t$, we shall use the property $$\mathbf{v}_{t_0}\cdot \mathbf{v_\perp}_{t} = 0 ^1\\\\\\\ \implies 10\cdot 10 \mathtt{\hat{i}}\cdot \mathtt{\hat{i}} + \{\ldots\}\mathtt{\hat {i}}\cdot \mathtt{\hat{j}} + (20 - t_0)\cdot (20-t)\mathtt{\hat{j}}\cdot\mathtt{\hat{j}} = 0 \implies (20-t_0)t=100 + 400 -20t_0 $$ . Now, only need to know what $t_0$ is.
$^1$I am assuming you know the relation, but if you don't then here is the little proof:
$\mathbf{A}\cdot \mathbf{B} = AB\cos\theta \implies \mathbf{A}\cdot \mathbf{B_{\perp A}} = AB \cos\left(\frac{\pi}{2}\right) = 0$