An object moving with uniform acceleration has a velocity of 12.0 cm/s in the positive x direction when its x coordinate is 3.00 cm if its x coordinate 2.00 s later is -5.00cm what is its acceleration

This question can be solved using the equations of motion.

The acceleration of the object is - 16 cm/s².

We can use the second equation of motion to find the acceleration of the object.

where,

Δs = distance covered = change in position = - 5 cm - 3 cm = - 8 cm

= initial velocity = 12 cm/s

t = time = 2 s

a = acceleration = ?

Therefore,

a = - 16 cm/s² (negative sign shows deceleration)

The attached picture shows the equations of motion.

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Given:

at time t=0, X0=X(0)=3.00 cm, v(0)=12 cm/s

at time t=2, x(2)=-5

Use kinematics equation:

X(t)=X(0)+v(0)*t+at^2/2

Substitute =>

X(t)=3+12t+at^2/2

At t=2, X(2)=-5 =>

-5=3+12(2)+a(2^2)/2 =>

acceleration = a =(-5-3-24)/2 = -16 cm/s^2

are there options? cause you ain’t. real specific