Answer :
Answer:
18.75m/s
Explanation:
Given data
initial speed ,u= o m/s
final speed v= 25 m/s
time t= 3.11 seconds
Applying the first equation of motion we have
[tex]v= u+at\\\\v-u=at\\\\a= \frac{v-u}{t} \\\\a= \frac{25-0}{6.22} \\\\a= 4.02 m/s^2[/tex]
Also applying [tex]v=u+at[/tex] with initial speed set at 0, we can also find speed at 3.11 seconds
[tex]v=0+4.02*3.11\\v= 12.50 m/s[/tex]
Hence the average speed in the first 3.11 seconds is
[tex]\bar v= \frac{v}{2} \\\\\bar v=\frac{ 12.50}{2} \\\\ \bar v= 6.25 m/s[/tex]
Applying the expression [tex]\bar v=\bar u+at[/tex] we can now fing the average speed in the second 3.11 second
[tex]\bar v=\bar u+at\\\\ \bar v=6.25+4.02*3.11\\\\ \bar v=18.75m/s[/tex]