Question

a student releases a marble from the top of a 190cm long ramp. the marble increases speed steadily and reaches the bottom of the ramp with a speed of 90.0cm/s. determine all unknowns and answer the following question. how long did the marble take to reach the bottom of the ramp? yellow box green box with s

Explanation:

Step1: Identify the formula

The marble is moving with constant acceleration (since it increases speed steadily), and we can use the formula for average velocity. The average velocity \( v_{avg} \) is given by \( v_{avg}=\frac{v_i + v_f}{2} \), where \( v_i \) is the initial velocity and \( v_f \) is the final velocity. Also, the distance \( d \) is related to average velocity and time \( t \) by \( d = v_{avg} \times t \). Since the marble is released from rest, \( v_i = 0 \, \text{cm/s} \), \( v_f=90.0 \, \text{cm/s} \), and \( d = 190 \, \text{cm} \).

First, calculate the average velocity:
\( v_{avg}=\frac{0 + 90.0}{2}=45.0 \, \text{cm/s} \)

Step2: Solve for time

Using the formula \( d = v_{avg} \times t \), we can rearrange for \( t \):
\( t=\frac{d}{v_{avg}} \)

Substitute \( d = 190 \, \text{cm} \) and \( v_{avg}=45.0 \, \text{cm/s} \):
\( t=\frac{190}{45.0}\approx4.22 \, \text{s} \) (rounded to two decimal places)

Answer:

\( \approx 4.22 \)