Question

A particle's position coordinates (x, y) are (3.0 m, 5.0 m) at t = 0; (6.0 m, 9.0 m) at t = 2.0 s; and (15.0 m, 15.0 m) at t = 5.0 s.

1.) Find the magnitude of the average velocity from t = 0 to t = 2 s.
=> ________m/2

2.) Find the magnitude of the average velocity from t = 0 to t = 5 s.
=> ________m/s

Answer 1

@theEric

Answer 2

The average velocity can be calculated with,
\[\vec{v}_m=\frac{\vec{r}_f-\vec{r}_i}{\Delta t}=\frac{(6,9)-(3,5)}{2}=(1.5,2.0)\ m/s\]
With magnitude,
\[v_m=\sqrt{1.5^2+2^2}=2.5\ m/s\]

For the secondo question,
\[\vec{v}_m=\frac{\vec{r}_f-\vec{r}_i}{\Delta t}=\frac{(15,15)-(3,5)}{2}=(6.0,5.0)\ m/s\]
\[v_m=\sqrt{5^2+6^2}=7.8\ m/s\]

Answer 3

The average velocity can be calculated with,
\[\vec{v}_m=\frac{\vec{r}_f-\vec{r}_i}{\Delta t}=\frac{(6,9)-(3,5)}{2}=(1.5,2.0)\ m/s\]
With magnitude,
\[v_m=\sqrt{1.5^2+2^2}=2.5\ m/s\]

For the secondo question,
\[\vec{v}_m=\frac{\vec{r}_f-\vec{r}_i}{\Delta t}=\frac{(15,15)-(3,5)}{2}=(6.0,5.0)\ m/s\]
\[v_m=\sqrt{5^2+6^2}=7.8\ m/s\]

Answer 4

#2 was apparently wrong. it's not 7.8 m/s I got the same answer... :/
@John_ES