A fish swimming at a constant speed of 0.5 m/s suddenly notices a shark appear behind it. 5 seconds later, the fish swimming in t

Question

A fish swimming at a constant speed of 0.5 m/s suddenly notices a shark appear behind it. 5 seconds later, the fish swimming in the same direction at a speed of 2.5 m/s. Calculate the fish’s acceleration. *

Reagan · Answer

Given     :       A fish swimming at a constant speed of 0.5 m/s [ as , Initial Velocity  (     u     )     ]  , After noticing the shark fish's speed is 2.5 m/s  [ as , Final Velocity (  v  )  ]  & Time taken is  5    seconds [ as , Time (  t    ) ]     Exigency     To     Find        :    The Acceleration of fish ?     ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀    ⠀⠀⠀⠀Given that ,     ⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀The initial velocity (  u     )  of fish is  0.5 m/s  .  ⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀The final velocity  (  v     )  of fish is  2.5 m/s    ⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀Total Time taken (  t    ) is   5 seconds      ⠀⠀⠀⠀⠀⠀⠀Finding   Acceleration     of fish :      dag : : pmb{ As, :We :know :that ::}      qquad  bigstar  : :  bf Acceleration  : :  :  sf The  : rate  :of  : of  : change  :of : velocity :is  :known  : as  : Acceleration  :.  :      qquad maltese : : bf Formula  :for  : Acceleration ::              qquad  dag : : bigg lgroup  pmb{ bf{ ; : Acceleration :(a) := : dfrac{  : : v  : :-  : :u : :}{t} : : }} bigg rgroup          ⠀⠀⠀⠀Here ,   u       is the Initial velocity,   v    is the final velocity  &   t       is the time taken .       qquad  dashrightarrow  sf Acceleration :(a) := : dfrac{  : : v  : :-  : :u : :}{t} qquad  :         ⠀⠀⠀⠀⠀⠀  underline { boldsymbol{ star :Now  : By  : Substituting  : the  : known  : Values  ::}}         qquad  dashrightarrow  sf Acceleration :(a) := : dfrac{  : : v  : :-  : :u : :}{t} qquad  :           qquad  dashrightarrow  sf Acceleration :(a) := : dfrac{  : : 2.5  : :-  : :0.5 : :}{5} qquad  :           qquad  dashrightarrow  sf Acceleration :(a) := : dfrac{  : : 2 : :}{5} qquad  :           qquad  dashrightarrow  sf Acceleration :(a) := : cancel { dfrac{  : : 2 : :}{5}} qquad  :           qquad  dashrightarrow  sf Acceleration :(a) := :0.4  : qquad  :           qquad  therefore  pmb{ underline{ purple{ frak{  :Acceleration :(a) := :0.4  :m/s^2   }}} } : : bigstar          therefore  : underline {  sf  Hence ,  : : The  :Acceleration  :of :fish :is  :  bf 0.4  : m/s^2  :}.

Ivy · Answer

Answer:   The acceleration of the fish is 0.4 m/s².    Explanation:     Fish was swimming at the speed of = 0.5 m/s   Time = 5 seconds  Fish's speed after noticing the shark = 2.5 m/s     Acceleration of the fish = ??    _____________________________     bigstar  : { boxed{ sf{ Acceleration =  dfrac{v - u}{t}}}}    v = 2.5 m/s  u = 0.5 m/s  t = 5 seconds        sf{ longrightarrow}  :   dfrac{2.5 - 0.5}{5}      sf{ longrightarrow}  :   dfrac{2}{5}      sf{ longrightarrow}  : 0.4  : {ms}^{ - 2}    Acceleration of the fish = 0.4 m/s²    Therefore,  the acceleration of the fish is 0.4 m/s².

Eva

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