A particle at *t*_{1}=−5.4 s is at *x*_{1}=3.4 cm and at *t*_{2}=5.6 s is at *x*_{2}=8.0 cm.

(a) What is its average velocity?

(b) Can you calculate its average speed from these data?

Solution: A particle at t1=−5.4 s is at x1=3.4 cm and at t2=5.6 s is at x2=8.0 cm.(a) What is its average velocity?(b) Can you calculate its average speed from these data?

A particle at *t*_{1}=−5.4 s is at *x*_{1}=3.4 cm and at *t*_{2}=5.6 s is at *x*_{2}=8.0 cm.

(a) What is its average velocity?

(b) Can you calculate its average speed from these data?

We're asked to calculate the average velocity of a particle given a change in position and a change in time.

Recall that the average velocity of an object, ${\stackrel{\rightharpoonup}{v}}_{avg}$, is related to the change in position (displacement,${\u2206}\stackrel{\rightharpoonup}{x}$) over change in time Δ*t*:

$\overline{){\stackrel{\mathbf{\rightharpoonup}}{\mathit{v}}}_{\mathit{a}\mathit{v}\mathit{g}}{\mathbf{=}}\frac{\mathbf{\u2206}\stackrel{\mathbf{\rightharpoonup}}{\mathit{x}}}{\mathbf{\u2206}\mathit{t}}{\mathbf{=}}\frac{{\mathit{x}}_{\mathbf{2}}\mathbf{-}{\mathit{x}}_{\mathbf{1}}}{{\mathit{t}}_{\mathbf{2}}\mathbf{-}{\mathit{t}}_{\mathbf{1}}}}$

Velocity is a **vector**, which means the **signs **really matter (unlike speed, which is a *scalar*.)

(a) For the particle in this problem, we're first asked to calculate its average velocity between *t*_{1} and *t*_{2}. We're given that *x*_{1} = 3.4 cm, *x*_{2} = 8.0 cm, *t*_{1} = −5.4 s, and *t*_{2} = 5.6 s.