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A skateboarder, starting at rest, rolls down a 10.1 m ramp. When she arrived at the bottom of the ramp her speed is 7.57 m/s. A) determine the magnitude of her acceleration, assumed to be constant. B) if the ramp is inclined at 26.0 degrees with respect to the ground, what is the component of her acceleration that is parallel to the ground?

A skateboarder starting at rest rolls down a 101 m ramp When she arrived at the bottom of the ramp her speed is 757 ms A determine the magnitude of her accelera class=

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Answer:

a) Acceleration = 2.84 m/s²

b) The component of her acceleration that is parallel to the ground = 2.55 m/s²

Explanation:

The diagram representing the illustration is drawn below:

a) Calculate the acceleration using the equation of motion below

[tex]\begin{gathered} v^2=u^2+2as \\ \\ 7.57^2=0^2+2(10.1)a \\ \\ 57.3049=20.2a \\ \\ a=\frac{57.3049}{20.2} \\ \\ a=2.84\text{ m/s}^2 \end{gathered}[/tex]

b) The component of here accleration that is parallel to the ground is:

[tex]\begin{gathered} a_x=acos\theta \\ \\ a_x=2.84cos26 \\ \\ a_x=2.55\text{ m/s}^2 \end{gathered}[/tex]

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