Respuesta :
Answers:
Final velocity = 29 m/s
Distance traveled = 73.5 meters
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Work Shown:
The given data is
[tex]v_i = 20 \text{ m}/\text{s} = \text{initial velocity}\\\\a = 3 \text{ m}/\text{s}^2 = \text{acceleration}\\\\t = 3 \text{ s} = \text{time duration}\\\\[/tex]
This leads to
[tex]v_f = v_i + a*t\\\\v_f = 20 + 3*3\\\\v_f = 20 + 9\\\\v_f = 29\\\\[/tex]
The final velocity is 29 m/s.
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We can use that final velocity to find the distance traveled.
[tex]x = 0.5*(v_i+v_f)*t\\\\x = 0.5*(20+29)*3\\\\x = 0.5*(49)*3\\\\x = 24.5*3\\\\x = 73.5\\\\[/tex]
The distance traveled is 73.5 meters.
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An alternative way to calculate the distance is to do this
[tex]x = \frac{(v_f)^2 - (v_i)^2}{2a}\\\\x = \frac{(29)^2 - (20)^2}{2*3}\\\\x = \frac{841 - 400}{6}\\\\x = \frac{441}{6}\\\\x = 73.5\\\\[/tex]
Or you could do this
[tex]x = v_i*t + 0.5*a*t^2\\\\x = 20*3 + 0.5*3*3^2\\\\x = 20*3 + 0.5*3*9\\\\x = 60 + 13.5\\\\x = 73.5\\\\[/tex]
For more information, check out the Kinematics equations.
