An object moving with uniform acceleration has a velocity of 15.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 1.60 s later is -5.00 cm, what is its acceleration?
To find the acceleration of the object, we can use the kinematic equation:
\[x = x_0 + v_0t + \frac{1}{2}at^2\]
Where:
- \(x\) is the final position of the object
- \(x_0\) is the initial position of the object
- \(v_0\) is the initial velocity of the object
- \(t\) is the time
- \(a\) is the acceleration
In this case, we are given:
Initial position, \(x_0 = 3.00 \, \text{cm}\)
Initial velocity, \(v_0 = 15.0 \, \text{cm/s}\)
Time, \(t = 1.60 \, \text{s}\)
Final position, \(x = -5.00 \, \text{cm}\)
Substituting these values into the equation, we have:
\[-5.00 \, \text{cm} = 3.00 \, \text{cm} + 15.0 \, \text{cm/s} \times 1.60 \, \text{s} + \frac{1}{2}a \times (1.60 \, \text{s})^2\]
Simplifying the equation:
\[-5.00 \, \text{cm} - 3.00 \, \text{cm} = 24.0 \, \text{cm} + 0.8a\]
\[-8.00 \, \text{cm} = 24.0 \, \text{cm} + 0.8a\]
Subtracting \(24.0 \, \text{cm}\) from both sides:
\[-32.0 \, \text{cm} = 0.8a\]
To solve for \(a\), we divide both sides by \(0.8 \, \text{cm/s}^2\):
\[\frac{-32.0 \, \text{cm}}{0.8 \, \text{cm/s}^2} = a\]
Simplifying, we find:
\[a = -40.0 \, \text{cm/s}^2\]
Therefore, the acceleration of the object is \(-40.0 \, \text{cm/s}^2\).