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?

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To find the acceleration of the object, we can use the kinematic equation for position with uniform acceleration:

x = x0 + v0*t + (1/2)*a*t^2

where:
x is the final position
x0 is the initial position
v0 is the initial velocity
t is the time
a is the acceleration

In this problem, we are given:
Initial position, x0 = 2.97 cm
Initial velocity, v0 = 11.0 cm/s
Final position, x = -5.00 cm
Time, t = 2.95 s

Plugging in the given values into the kinematic equation, we have:

-5.00 cm = 2.97 cm + (11.0 cm/s)*(2.95 s) + (1/2)*a*(2.95 s)^2

Let's solve this equation for the acceleration, a.

First, simplify the equation:

-5.00 cm - 2.97 cm = (11.0 cm/s)*(2.95 s) + (1/2)*a*(2.95 s)^2

-7.97 cm = 32.45 cm + (1/2)*a*(8.7025 s^2)

Next, isolate the acceleration term:

-40.42 cm = a*(4.35125 s^2)

Finally, solve for the acceleration:

a = (-40.42 cm) / (4.35125 s^2)
a ≈ -9.30 cm/s^2

Therefore, the acceleration of the object is approximately -9.30 cm/s^2.