A rabbit is moving in the positive x-direction at 2.70 m/s when it spots a predator and accelerates to a velocity of 13.3 m/s along the positive y-axis, all in 2.40 s. Determine the x-component and the y-component of the rabbit's acceleration. (Enter your answers in m/s2. Indicate the direction with the signs of your answers.)

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Answer:

[tex]a_x=-1.125\ m.s^{-2}[/tex]

[tex]a_y=5.5416\ m.s^{-2}[/tex]

Explanation:

Given:

  • velocity of rabbit in positive x direction, [tex]v_x=2.7\ m.s^{-1}[/tex]
  • velocity of rabbit in positive y direction, [tex]v_y=13.3\ m.s^{-1}[/tex]
  • time taken to change the direction of velocity, [tex]t=2.4\ s[/tex]

When the velocity is in x direction then the y component is zero and vice-versa.

The velocity in x direction goes from [tex]2.7\ m.s^{-1} to\ 0\ m.s^{-1}[/tex], therefore acceleration:

[tex]a_x=\frac{0-2.7}{2.4}[/tex]

[tex]a_x=-1.125\ m.s^{-2}[/tex]

The velocity in y direction goes from [tex]0\ m.s^{-1}\ to\ 13.3\ m.s^{-1}[/tex], therefore acceleration:

[tex]a_y=\frac{13.3-0}{2.4}[/tex]

[tex]a_y=5.5416\ m.s^{-2}[/tex]

Acceleration is defined as the rate of changing the velocity of the body. The acceleration in the positive x-direction is -1.125 m/sec².While the acceleration in the positive y-direction is 5.541 m/sec².

What is acceleration?

Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².It is a vector quantity. It requires both magnitudes as well as direction to define.

The given data in the problem is

[tex]\rm v_x= 2.70 m/sec}[/tex]

[tex]\rm v_y=13.3 m/sec}[/tex]

t = 2.40 sec.

The rat moves in one direction at a time

The rat moves from the velocity of 2.7 m/sec to rest. The acceleration in the positive x-direction will be

[tex]\rm a_x = \frac{v_x-u_x}{t}} \\\\a_x = \frac{0-2.7}{2.4}\\\\a_x = -1.125 \;\rm {m/sec^2}[/tex]

Hence the acceleration in the positive x-direction is -1.125 m/sec².

The rat moves from rest to the velocity of 13.3m/sec. The acceleration in the positive x-direction will be

[tex]\rm a_y = \frac{v_y-u_y}{t}} \\\\a_y = \frac{13.3-0}{2.4}\\\\a_y= 5.541 \;\rm {m/sec^2}[/tex]

Hence the acceleration in the positive y-direction is 5.541 m/sec².

To learn more about acceleration refer to the link;

https://brainly.com/question/969842

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