Answer :
In an uniform accelerated motion, the distance covered by the fish is given by:
[tex]S= \frac{1}{2}at^2 [/tex]
where
a is the acceleration
t is the time
The acceleration is equal to the increase in speed of the fish divided by the time taken:
[tex]a= \frac{v_f-v_i}{t}= \frac{4 m/s-0m/s}{0.11 s}=36.4 m/s^2 [/tex]
If substitute the acceleration into the first equation, we find the distance covered by the fish:
[tex]S= \frac{1}{2}at^2= \frac{1}{2}(36.4 m/s^2)(0.11 s)^2=0.22 m [/tex]
[tex]S= \frac{1}{2}at^2 [/tex]
where
a is the acceleration
t is the time
The acceleration is equal to the increase in speed of the fish divided by the time taken:
[tex]a= \frac{v_f-v_i}{t}= \frac{4 m/s-0m/s}{0.11 s}=36.4 m/s^2 [/tex]
If substitute the acceleration into the first equation, we find the distance covered by the fish:
[tex]S= \frac{1}{2}at^2= \frac{1}{2}(36.4 m/s^2)(0.11 s)^2=0.22 m [/tex]
