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. *

Answer 1

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.

Answer 2

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 .

$\qquad \dashrightarrow \sf Acceleration\:(a)\:=\:\dfrac{ \:\: v \:\:- \:\:u\:\:}{t}\qquad \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf Acceleration\:(a)\:=\:\dfrac{ \:\: v \:\:- \:\:u\:\:}{t}\qquad \:$

$\qquad \dashrightarrow \sf Acceleration\:(a)\:=\:\dfrac{ \:\: 2.5 \:\:- \:\:0.5\:\:}{5}\qquad \:$

$\qquad \dashrightarrow \sf Acceleration\:(a)\:=\:\dfrac{ \:\: 2\:\:}{5}\qquad \:$

$\qquad \dashrightarrow \sf Acceleration\:(a)\:=\:\cancel {\dfrac{ \:\: 2\:\:}{5}}\qquad \:$

$\qquad \dashrightarrow \sf Acceleration\:(a)\:=\:0.4 \:\qquad \:$

$\qquad \therefore \pmb{\underline{\purple{\frak{ \:Acceleration\:(a)\:=\:0.4 \:m/s^2 }}} }\:\:\bigstar$

$\therefore \:\underline { \sf Hence , \:\: The \:Acceleration \:of\:fish\:is \: \bf 0.4 \: m/s^2 \:}.$