A body moving with uniform acceleration has a velocity of 11.1cm/s when it's coordinate is 3.59cm. If it's x coordinate 2.48sec later is -3.58cm, what is the x-component of it's acceleration. Answer in units of cm/s^2
counting from t=0 when x=3.59,
x(t) = 3.59 + 11t + 1/2 at^2
-3.58 = 3.59 + 11*2.48 + 1/2 a * 2.48^2
a = -11.2 cm/s^2
To find the x-component of the acceleration, we can use the kinematic equation:
\(v^2 = u^2 + 2a \cdot s\)
where:
- \(v\) is the final velocity
- \(u\) is the initial velocity
- \(a\) is the acceleration
- \(s\) is the displacement
Given:
\(u = 11.1 \, \text{cm/s}\)
\(v = -3.58 \, \text{cm/s}\)
\(s = -3.58 - 3.59 = -7.17 \, \text{cm}\)
Substituting these values into the equation, we have:
\((-3.58)^2 = (11.1)^2 + 2 \cdot a \cdot (-7.17)\)
Simplifying:
\(12.8164 = 123.21 + (-14.34a)\)
Rearranging the equation:
\(-14.34a = 12.8164 - 123.21\)
\(-14.34a = -110.3936\)
Finally, solving for \(a\):
\(a = \frac{-110.3936}{-14.34}\)
\(a \approx 7.694 \, \text{cm/s}^2\)
Therefore, the x-component of the acceleration is approximately \(7.694 \, \text{cm/s}^2\).