Average Velocity and Instantaneous Velocity — Question 3

Question 3

Differentiate each position component

A particle has $x(t)=(2\ \mathrm{m/s})t$ and $y(t)=(1\ \mathrm{m/s^2})t^2$. Instantaneous velocity is $\vec v=d\vec r/dt$ and speed is $|\vec v|$.

1. Find the velocity and speed at 3 s.

2. Find the average velocity from 1 s to 3 s. At what time in that interval does instantaneous velocity equal this average?

3. Find the slope of the path at 3 s and connect it with the velocity direction.

Question 3 - Solution

Differentiate each position component

(a) Differentiate before evaluating the time.

$$\vec v=\frac{d\vec r}{dt},\quad\vec v(3)=(2,6)\,\mathrm{m/s},\quad\boxed{v(3)=\sqrt{40}\,\mathrm{m/s}}$$ $v_x=2$ m/s and $v_y=(2\ \mathrm{m/s^2})t$. At 3 s, $\vec v=(2,6)$ m/s and speed is $\sqrt{40}=2\sqrt{10}\approx6.325$ m/s. Direction is $\tan^{-1}(3)\approx71.565^\circ$ above positive x.

(b) Compare a chord rate with a local rate.

$$\vec v_{\mathrm{av}}=\frac{(6,9)-(2,1)}{3-1}=(2,4)\,\mathrm{m/s}=\vec v(2\,\mathrm s)$$ Positions at 1 and 3 s are $(2,1)$ and $(6,9)$ m. Thus the average is $(2,4)$ m/s, equal to instantaneous velocity at 2 s. The equality at the midpoint follows here from velocity being linear in time; it is not universal.

(c) Use component ratios for a tangent.

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{6}{2}=3,\quad\theta=\tan^{-1}3=71.565^\circ$$ $dy/dx=(dy/dt)/(dx/dt)=v_y/v_x=3$ at 3 s. The velocity arrow is tangent to the curve and points along increasing time. Its magnitude supplies the rate of motion, which a tangent line alone does not specify.