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A particle travels horizontally between two parallel walls separated by 18.4 m. It moves toward the opposing wall at a constant rate of 9.5 m/s. Also, it has an acceleration in the direction parallel to the walls of 3.7 m/s 2 . 18.4 m 3.7 m/s 2 9.5 m/s What will be its speed when it hits the opposing wall? Answer in units of m/s

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lublana

Answer:

7.18m/s

Explanation:

We are given that

Distance,s=18.4 m

Speed,v=9.5 m/s

Acceleration,a=[tex]3.7m/s^2[/tex]

We have to find the speed when it hits the opposing wall.

In perpendicular direction,a=0

Time,[tex]t=\frac{distance}{speed}=\frac{18.4}{9.5}=1.94 s[/tex]

In parallel direction

Initial velocity,u=0

[tex]v=u+at[/tex]

Substitute the values

[tex]v=0+3.7(1.94)=7.18 m/s[/tex]

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