1) Average velocity: 19 mi/h
The average velocity is given by:
[tex]v=\frac{d}{t}[/tex]
where
d is the displacement
t is the time taken
In this problem,
d = 3.8 mi is the displacement
[tex]t=12 min \cdot \frac{1}{60 min/h}=0.2 h[/tex] is the time taken
Substituting,
[tex]v=\frac{3.8 mi}{0.2 h}=19 mi/h[/tex]
Note that this is actually the average speed, not the average velocity, since there is no information about the direction of the motion.
2) [tex]10,309 mi/h^2[/tex]
The acceleration is given by
[tex]a=\frac{v-u}{t}[/tex]
where we have
v = 28 mph is the final velocity
u = 18 mph is the initial velocity
t = 3.5 s is the time taken
Conerting the time from seconds to hours,
[tex]t=3.5 s \cdot \frac{1}{3600 s/h}=9.7\cdot 10^{-4} h[/tex]
So the acceleration is
[tex]a=\frac{28 mph-18 mph}{9.7\cdot 10^{-4} h}=10,309 mi/h^2[/tex]
