Question

Q1: A particle moves from x = 3.0 m to x = 9.0 m in 3.4 s, reaching a final speed of 3.00 m/s once it reaches x = 9.0 m. What is the (uniform) acceleration during this interval?

Answer 1

The equation of the particle movement with the constant acceleration is:

$$\Delta x = v_0t + \dfrac{at^2}{2}$$

where $\Delta x = 9m-3m = 6m$ is the trevelled distance, $v_0$ is initial speed, $t = 3.4s$ is the time and $a$ is the acceleration.

The equation for velocity is:

$$v = v_0 + at$$

where $v = 3 m/s$ is the final velocity of the particle.

Solving this two equations simultaneously, obtain:

$$v_0 = v - at\\ \Delta x = (v-at)t + \dfrac{at^2}{2} = vt - at^2 + \dfrac{at^2}{2} = vt-\dfrac{at^2}{2}$$

Expressing $a$, get:

$$a = \dfrac{2v}{t} - \dfrac{2\Delta x}{t^2} = \dfrac{2\cdot 3}{3.4} - \dfrac{2\cdot 6}{3.4^2} \approx 0.73m/s^2$$

Answer. 0.73 m/s^2.