An object moving with uniform acceleration has a velocity of 11.0 cm/s in the positive x-direction when its x-coordinate is 2.97 cm. If its x-coordinate 2.95 s later is −5.00 cm, what is its acceleration?
To find the acceleration of an object moving with uniform acceleration, we can use the kinematic equation for displacement:
x = x₁ + v₁t + (1/2)at²
where:
x is the final position
x₁ is the initial position
v₁ is the initial velocity
t is the time
a is the acceleration
In this case, we are given the following values:
x₁ = 2.97 cm (initial position)
v₁ = 11.0 cm/s (initial velocity)
t = 2.95 s (time)
x = -5.00 cm (final position)
We need to solve for the acceleration, a. Rearranging the equation, we get:
a = (2x - 2x₁ - v₁t) / t²
Substituting the given values, we have:
a = (2*(-5.00 cm) - 2*(2.97 cm) - (11.0 cm/s)*(2.95 s)) / (2.95 s)²
Now, we can simplify and calculate the acceleration:
a = (-10.0 cm - 5.94 cm - 32.45 cm) / (2.95 s)²
a = (-48.39 cm) / (2.95 s)²
a ≈ -6.56 cm/s²
Therefore, the acceleration of the object is approximately -6.56 cm/s² in the negative x-direction.