Answer:
[tex]56.5\ \text{s}[/tex]
[tex]21.13\ \text{m/s}[/tex]
Explanation:
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time
s = Displacement
Here the kinematic equations of motion are used
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}[/tex]
Time the car is at constant velocity is 39 s
Time the car is decelerating is 5 s
Total time the car is in motion is [tex]12.5+39+5=56.5\ \text{s}[/tex]
Distance traveled
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}[/tex]
[tex]s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}[/tex]
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}[/tex]
The total displacement of the car is [tex]156.25+975+62.5=1193.75\ \text{m}[/tex]
Average velocity is given by
[tex]\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}[/tex]
The average velocity of the car is [tex]21.13\ \text{m/s}[/tex].
