An object moving with uniform acceleration has a velocity of 10.0 cm/s in the positive x-direction when its x-coordinate is 2.91 cm. If its x-coordinate 1.85 s later is −5.00 cm, what is its acceleration?
answered at 12:01
To solve this problem, we can use the equations of motion for an object moving with uniform acceleration. Let's break down the given information:
Initial velocity (u) = 10.0 cm/s
Initial position (x1) = 2.91 cm
Final position (x2) = -5.00 cm
Time interval (t) = 1.85 s
Using the equation:
x2 = x1 + ut + (1/2)at^2
We can substitute the given values and solve for acceleration (a).
-5.00 cm = 2.91 cm + (10.0 cm/s)(1.85 s) + (1/2)a(1.85 s)^2
Simplifying the equation:
-5.00 cm - 2.91 cm = 18.5 cm/s - (1/2)a(3.4225 s^2)
-7.09 cm = 18.5 cm/s - 1.71125a
Rearranging the equation:
1.71125a = 18.5 cm/s + 7.09 cm
a = (18.5 cm/s + 7.09 cm) / 1.71125
a = 25.59 cm/s / 1.71125
a ≈ 14.956 cm/s^2
Therefore, the acceleration of the object is approximately 14.956 cm/s^2.