Question

A particle moves in the x-y plane with the velocity $\vec{v} = a\hat{i} + bt\hat{j}$. At the instant $t = a\sqrt{3}/b$ the magnitude of tangential, normal and total acceleration are ______, ______, & ______.

Answer 1

The magnitude of tangential, normal and total acceleration are $\frac{b\sqrt{3}}{2}$, $\frac{b}{2}$, & $b$.

Answer 2

The velocity of the particle is given by $\vec{v} = a\hat{i} + bt\hat{j}$. The acceleration vector is the time derivative of velocity: $\vec{a} = \frac{d\vec{v}}{dt} = b\hat{j}$. The magnitude of the total acceleration is $|\vec{a}| = \sqrt{0^2 + b^2} = b$.

The speed of the particle is $v = |\vec{v}| = \sqrt{a^2 + (bt)^2} = \sqrt{a^2 + b^2t^2}$. The tangential acceleration is the rate of change of speed: $a_t = \frac{dv}{dt} = \frac{d}{dt}(a^2 + b^2t^2)^{1/2} = \frac{b^2t}{\sqrt{a^2 + b^2t^2}}$. The normal acceleration $a_n$ can be found using the relation $a^2 = a_t^2 + a_n^2$. $a_n = \sqrt{a^2 - a_t^2} = \sqrt{b^2 - \left(\frac{b^2t}{\sqrt{a^2 + b^2t^2}}\right)^2} = \sqrt{\frac{a^2b^2}{a^2 + b^2t^2}} = \frac{|ab|}{\sqrt{a^2 + b^2t^2}}$. Assuming $a>0$ and $b>0$, $a_n = \frac{ab}{\sqrt{a^2 + b^2t^2}}$.

At the instant $t = \frac{a\sqrt{3}}{b}$: $a^2 + b^2t^2 = a^2 + b^2\left(\frac{a\sqrt{3}}{b}\right)^2 = a^2 + 3a^2 = 4a^2$. So, $\sqrt{a^2 + b^2t^2} = \sqrt{4a^2} = 2a$ (assuming $a>0$).

Substituting this into the expressions for $a_t$, $a_n$, and $|\vec{a}|$: Tangential acceleration: $a_t = \frac{b^2(a\sqrt{3}/b)}{2a} = \frac{ab\sqrt{3}}{2a} = \frac{b\sqrt{3}}{2}$. Normal acceleration: $a_n = \frac{ab}{2a} = \frac{b}{2}$. Total acceleration: $|\vec{a}| = b$.

Thus, the magnitudes of tangential, normal, and total acceleration are $\frac{b\sqrt{3}}{2}$, $\frac{b}{2}$, and $b$, respectively.