Question

A particle undergoes a constant acceleration of 3.90 m/s2. After a certain amount of time, its velocity is 12.2 m/s. (Where applicable, indicate the direction with the sign of your answer.) (a) If its initial velocity is 6.1 m/s, what is its displacement during this time?

Answer 1

Answer:

$s_{f} = 14.312 m$

Explanation:

Since the particle is experimenting a constant acceleration, the displacement can be found by using this formula:

$v_{f}^{2} = v_{o}^{2} + 2 \cdot a \cdot (s_{f} - s_{o})$

Since $s_{o} = 0 m$, the equation is simplified to this form:

$v_{f}^{2} = v_{o}^{2} + 2 \cdot a \cdot s_{f}$

Then, the displacement is now isolated:

$s_{f} = \frac{v_{f}^{2}-v_{o}^{2}}{2 \cdot a}$

Terms are replaced herein:

$s_{f} = \frac{(12.2 \frac{m}{s})^{2}-(6.1 \frac{m}{s})^{2}}{2 \cdot (3.90 \frac{m}{s^{2}}) } \\s_{f} = 14.312 m$