Average Velocity and Instantaneous Velocity — Question 10

Question 10

Speed is not the rate of increase of distance from the origin

A particle has $\vec r(t)=[3\hat\imath+(4\ \mathrm{s^{-1}})t\hat\jmath]$ m. Define $\rho(t)=|\vec r(t)|$, its distance from the origin.

1. Find the velocity and speed at every time.

2. Find $d\rho/dt$ at 0 and at 1 s by differentiating $\rho(t)$.

3. Explain why the radial rate can be zero while speed is nonzero, and why the radial rate cannot exceed speed in magnitude.

Question 10 - Solution

Speed is not the rate of increase of distance from the origin

(a) Differentiate the position vector.

$$\boxed{\vec v=\frac{d(3,4u)}{dt}\,\mathrm m=(0,4)\,\mathrm{m/s}},\quad v=4\,\mathrm{m/s}$$ $\vec v=4\hat\jmath$ m/s, so the speed is always 4 m/s. The motion is a straight vertical line with x fixed at 3 m.

(b) Differentiate the scalar distance separately.

$$\rho=\sqrt{9+16u^2}\,\mathrm m,\quad\dot\rho=\frac{16u}{\sqrt{9+16u^2}}\,\mathrm{m/s}:\quad\dot\rho(0)=0,\ \dot\rho(1)=3.2\,\mathrm{m/s}$$ With $u=t/(1\ \mathrm s)$, $\rho=\sqrt{9+16u^2}$ m and $d\rho/dt=16u/\sqrt{9+16u^2}$ m/s. This is zero at 0 and $16/5=3.2$ m/s at 1 s. Neither value is the full speed.

(c) Interpret a projection of velocity.

$$\dot\rho=\frac{\vec r\cdot\vec v}{|\vec r|}=\hat r\cdot\vec v,\quad|\dot\rho|\le|\hat r||\vec v|=v$$ The radial rate is the velocity component along the outward radius. At 0, the radius is horizontal and velocity vertical, so their projection is zero. For $\rho>0$, $d\rho/dt=\hat r\cdot\vec v$; a projection has magnitude at most $|\vec v|$. Motion perpendicular to the radius still contributes to speed.