Question

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?

Answer 1

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

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

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.