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A speedy rabbit is hopping to the right with a velocity of 4.0m/s when it sees a carrot in the distance. The rabbit speeds up to its maximum velocity of 13m/s with a constant acceleration 2.0m/s2 rightward.

How many seconds does it take the rabbit to speed up from 4.0m/s to 13m/s

Respuesta :

Answer:

t = 4.5 seconds

Step-by-step explanation:

initial velocity = 4.0 m/s

final velocity = 13 m/s

acceleration = 2.0 m/s2

t = (final velocity - initial velocity) / acceleration

t = (13 - 4.0) / 2.0

t = 4.5 seconds

Answer:

4.5 sec.

Step-by-step explanation:

We know that:

[tex]v_{0}=4.0m/s[/tex]

[tex]v_{f}=13m/s[/tex]

[tex]a=2.0m/s^{2}[/tex]

Now, to calculate the time, we must use this equation:

[tex]v_{f}=v_{0}+at\\v_{f}-v_{0}=at\\\frac{v_{f}-v_{0}}{a}=t \\t=\frac{13m/s-4m/s}{2m/s^{2} } =4.5sec[/tex]

This means that the rabbit needs 4.5 seconds to speed up from 4m/s to 13 m/s. In this type of problem, we just need to read carefully if there are initial ([tex]v_{0}[/tex]) and final speed ([tex]v_{f}[/tex]), and acceleration, because that's the case when we use the equation showed.

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