An object moving with uniform acceleration has a velocity of 12m/s in the +ve x-direction,when its x-coordinate 3cm.if its x-coordinate 2sec. later is -5cm ,what is the acceleration?
-16cm/s^2
d=vi(t)+1/2 a t^2
-8=12*2+1/2 a (4)
solve for a
i went physics qustion answer
To find the acceleration of the object, we can use the equation of motion:
\[x = x_0 + v_0 t + \frac{1}{2} a t^2\]
where:
- \(x\) is the final position
- \(x_0\) is the initial position
- \(v_0\) is the initial velocity
- \(t\) is the time elapsed
- \(a\) is the acceleration
Given the following information:
- \(x_0 = 3\) cm = 0.03 m (initial position)
- \(v_0 = 12\) m/s (initial velocity)
- \(t = 2\) sec (time elapsed)
- \(x = -5\) cm = -0.05 m (final position)
We can rewrite the equation as:
\[-0.05 = 0.03 + 12(2) + \frac{1}{2} a(2)^2\]
Simplifying the equation, we get:
\[0 = 0.03 + 24 + 2a\]
Rearranging the equation to isolate \(a\), we have:
\[2a = -0.03 - 24\]
\[2a = -24.03\]
\[a = \frac{-24.03}{2}\]
Thus, the acceleration of the object is approximately \(-12.015\) m/s².