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A car driving at 20m/s accelerates at 3m/s2 for 3s. What is its final velocity? How far will it travel during this time?

User Zulan
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1 Answer

8 votes

Answers:

Final velocity = 29 m/s

Distance traveled = 73.5 meters

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Work Shown:

The given data is


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}\\\\

This leads to


v_f = v_i + a*t\\\\v_f = 20 + 3*3\\\\v_f = 20 + 9\\\\v_f = 29\\\\

The final velocity is 29 m/s.

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We can use that final velocity to find the distance traveled.


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\\\\

The distance traveled is 73.5 meters.

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An alternative way to calculate the distance is to do this


x = ((v_f)^2 - (v_i)^2)/(2a)\\\\x = ((29)^2 - (20)^2)/(2*3)\\\\x = (841 - 400)/(6)\\\\x = (441)/(6)\\\\x = 73.5\\\\

Or you could do this


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\\\\

For more information, check out the Kinematics equations.

User AreToo
by
5.9k points