Question

a student releases a marble from the top of a 220cm long ramp. the marble increases speed steadily and reaches the bottom of the ramp in 3.70s. what was the magnitude of the marbles acceleration? what was the marbles final speed?

Explanation:

Step1: Convert the length of the ramp to meters

The length of the ramp \(s = 220\space cm=2.20\space m\), initial velocity \(u = 0\space m/s\) (released from rest), time \(t = 3.70\space s\).
Use the second - equation of motion \(s=ut+\frac{1}{2}at^{2}\). Since \(u = 0\), the equation simplifies to \(s=\frac{1}{2}at^{2}\).

Step2: Solve for acceleration \(a\)

Rearrange \(s=\frac{1}{2}at^{2}\) for \(a\). We get \(a=\frac{2s}{t^{2}}\).
Substitute \(s = 2.20\space m\) and \(t = 3.70\space s\) into the formula:
\(a=\frac{2\times2.20}{3.70^{2}}=\frac{4.40}{13.69}\approx0.321\space m/s^{2}\)

Step3: Solve for final speed \(v\)

Use the first - equation of motion \(v = u+at\). Since \(u = 0\), \(v=at\).
Substitute \(a\approx0.321\space m/s^{2}\) and \(t = 3.70\space s\) into the formula:
\(v=0.321\times3.70 = 1.19\space m/s\)

Answer:

• Acceleration: \(0.321\space m/s^{2}\)
• Final speed: \(1.19\space m/s\)