Lesson Plan

Calculate the average speed of an object as distance/time, and rearrange the relation to solve for distance or time

Distinguish instantaneous speed from average speed

Calculate average velocity from displacement and time using v_avg = Δd/Δt

Distinguish speed (a scalar) from velocity (a vector), and explain why the magnitude of average velocity is not generally equal to average speed

Distinguish instantaneous velocity from average velocity

2.2

"Speed is the rate at which an object changes its location. Like distance, speed is a scalar because it has a magnitude but not a direction." "$\Delta t = t_f - t_0.$" "Average speed, vavg, is the distance traveled divided by the time during which the motion occurs." "Its speed at a specific instant in time, however, is its instantaneous speed. A car's speedometer describes its instantaneous speed." "Velocity describes the speed and direction of an object. ... Average velocity is displacement divided by the time over which the displacement occurs." "$\vec{v}{\text{avg}} = \frac{\text{displacement}}{\text{time}} = \frac{\Delta\vec{d}}{\Delta t} = \frac{\vec{d}f - \vec{d}0}{t_f - t_0}$" "If your car's odometer shows the total distance traveled was 6 km, then your average speed was 12 km/h. Your average velocity, however, was zero because your displacement for the round trip is zero." "A student has a displacement of 304 m north in 180 s. ... $\vec{v}{\text{avg}} = \frac{\Delta\vec{d}}{\Delta t} = \frac{304 \text{ m}}{180 \text{ s}} = 1.7 \text{ m/s north}$" "Layla jogs with an average velocity of 2.4 m/s east. ... $\Delta\vec{d} = \vec{v}{\text{avg}} , \Delta t = (2.4 \text{ m/s})(46 \text{ s}) = 1.1 \times 10^2 \text{ m east}$" "Phillip walks ... 428 m west with an average velocity of 1.7 m/s west? ... $\Delta t = \frac{\Delta\vec{d}}{\vec{v}_{\text{avg}}} = \frac{428 \text{ m}}{1.7 \text{ m/s}} = 2.5 \times 10^2 \text{ s}$"