An object moving with uniform acceleration has a velocity of 10.0 cm/s in the positive x-direction when its x-coordinate is 3.09 cm. If its x-coordinate 2.75 s later is −5.00 cm, what is its acceleration?
To find the acceleration of the object, we can use the kinematic equation:
\[x = x_0 + v_0t + \frac{1}{2}at^2\]
Where:
- x is the final x-coordinate (-5.00 cm)
- x₀ is the initial x-coordinate (3.09 cm)
- v₀ is the initial velocity (10.0 cm/s)
- t is the time interval (2.75 s)
- a is the acceleration (what we want to find)
First, let's rearrange the equation to solve for acceleration (a):
\[a = \frac{2(x - x_0 - v_0t)}{t^2}\]
Now we can substitute the known values into the equation:
\[a = \frac{2((-5.00) - (3.09) - (10.0)(2.75))}{(2.75)^2}\]
Simplifying this expression will give us the value of acceleration.