Question

a student releases a marble from the top of a ramp. the marble increases speed steadily and travels 80.0cm in 4.60s. what was the marbles final speed? cm/s

Explanation:

Step1: Use the kinematic equation for uniformly - accelerated motion

The kinematic equation for displacement \(s = v_{0}t+\frac{1}{2}at^{2}\). Since the marble is released from rest (\(v_{0} = 0\)), the equation simplifies to \(s=\frac{1}{2}at^{2}\). First, find the acceleration \(a\).

$$a=\frac{2s}{t^{2}}$$

Substitute \(s = 80.0\space cm\) and \(t = 4.60\space s\)

$$a=\frac{2\times80.0}{4.60^{2}}=\frac{160}{21.16}\approx7.56\space cm/s^{2}$$

Step2: Use the equation \(v = v_{0}+at\) to find the final speed

Since \(v_{0} = 0\), the equation for final speed \(v\) is \(v = at\)
Substitute \(a\approx7.56\space cm/s^{2}\) and \(t = 4.60\space s\)

$$v=7.56\times4.60 = 34.8\space cm/s$$

Another way:

Step1: Use the formula for average velocity in uniformly - accelerated motion

The average velocity \(\overline{v}=\frac{s}{t}\), and for uniformly - accelerated motion \(\overline{v}=\frac{v_{0} + v}{2}\). Since \(v_{0}=0\), \(\overline{v}=\frac{v}{2}\)
First, calculate the average velocity \(\overline{v}=\frac{s}{t}\)
Substitute \(s = 80.0\space cm\) and \(t = 4.60\space s\)

$$\overline{v}=\frac{80.0}{4.60}\approx17.4\space cm/s$$

Step2: Find the final speed \(v\)

Since \(\overline{v}=\frac{v}{2}\), then \(v = 2\overline{v}\)
Substitute \(\overline{v}\approx17.4\space cm/s\)

$$v=2\times17.4 = 34.8\space cm/s$$

Answer:

\(34.8\space cm/s\)