Question

An object moving with uniform acceleration has a velocity of 11.0 cm/s in the positive x-direction when its x-coordinate is 2.76 cm. If its x-coordinate 2.25 s later is −5.00 cm, what is its acceleration?

Answer 1

Given:

The initial velocity is

$v_i=\text{ 11 cm/s}$

The initial position is x1 = 2.76 cm

The final position is x2 = - 5 cm

The time will be t = 2.25 s

To find the acceleration.

Explanation:

The acceleration can be calculated by the formula

$\begin{gathered} \Delta x=v_it+\frac{1}{2}at^2 \\ \frac{1}{2}at^2=\Delta x-v_it \\ a=\frac{2(\Delta x-v_it)}{t^2} \end{gathered}$

Here, the displacement is

$\begin{gathered} \Delta x=x2-x1 \\ =-5-2.76 \\ =-7.76\text{ cm} \end{gathered}$

On substituting the values, the acceleration will be

$\begin{gathered} a=\frac{2(-7.76-11\times2.25)}{(2.25)^2} \\ =\text{ -12.84 cm/s}^2 \end{gathered}$

Final Answer: The acceleration is -12.84 cm/s^2