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A ladybug is moving at a speed of 1 cm/s. She begins to decelerate at a rate of –0.25 cm/s2 until she comes to rest. How much distance does the ladybug cover from the moment she begins to decelerate until the moment she comes to rest?

Sagot :

Answer:

Ladybug covers [tex]$s=2cm$[/tex] distance from the moment she begins to decelerate until she comes to the rest.

Explanation:

• The net work done on a body is equal to change in kinetic energy of the body. This is called as Work-energy theorem. Numerically,

[tex]${K_f} - {K_i} = W$[/tex]

• To find the distance, which Ladybug cover from the moment she begins to decelerate until she comes to the rest, use the formula: [tex]${v^2} = {u^2} + 2as$[/tex]

Where, [tex]$v$[/tex] is final velocity, [tex]$u$[/tex] is initial velocity, [tex]$a$[/tex] is acceleration and [tex]$s$[/tex] is displacement.

• After moving, ladybug comes to the rest, so final velocity, [tex]$v = 0cm/sec$[/tex].

• Placing the value of the given initial velocity,[tex]$u=1cm/s$[/tex] , acceleration, [tex]$a = - 0 \cdot 25cm/se{c^2}$[/tex] and final velocity, [tex]$v = 0$\\[/tex] in the above formula.

[tex]$\[\begin{align}& \therefore{v^2} = {u^2} + 2as \\& \Rightarrow 0 = {\left( 1 \right)^2} + 2\left( { - 0 \cdot 25} \right)\left( s \right) \\& \Rightarrow 0 = 1 - 0 \cdot 5s \\& \Rightarrow 0 \cdot 5s = 1 \\& \Rightarrow s = 2cm \\\end{align}\]$[/tex]

• Hence, Ladybug covers [tex]$s=2cm$[/tex] distance from the moment she begins to decelerate until she comes to the rest.

Learn more about velocity here:

https://brainly.com/question/20771493

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