Question

a student releases a marble from the top of a 80cm long ramp. the marble increases speed steadily and reaches the bottom of the ramp with a speed of 50cm/s. determine all unknowns and answer the following question.
$s_i = 0$ cm/s
$s = 50$ cm/s

Explanation:

Step1: Identify the motion type

The marble moves with constant acceleration (since speed increases steadily), initial velocity \( s_i = 0 \, \text{cm/s} \), final velocity \( s_f = 50 \, \text{cm/s} \), distance \( d = 80 \, \text{cm} \). We can use the equation \( v_f^2 = v_i^2 + 2ad \) to find acceleration \( a \), or \( d=\frac{(v_i + v_f)}{2}t \) to find time \( t \). Let's find time first.

Step2: Use the average velocity formula

Average velocity \( v_{avg}=\frac{v_i + v_f}{2}=\frac{0 + 50}{2}=25 \, \text{cm/s} \).

Step3: Find time using \( d = v_{avg} \times t \)

Rearrange for \( t \): \( t=\frac{d}{v_{avg}}=\frac{80}{25}=3.2 \, \text{s} \).

Step4: Find acceleration using \( v_f = v_i + at \)

Rearrange for \( a \): \( a=\frac{v_f - v_i}{t}=\frac{50 - 0}{3.2}=15.625 \, \text{cm/s}^2 \).

Answer:

• Initial velocity \( s_i = 0 \, \text{cm/s} \), final velocity \( s_f = 50 \, \text{cm/s} \), distance \( d = 80 \, \text{cm} \).
• Time to reach bottom: \( \boldsymbol{3.2 \, \text{seconds}} \), Acceleration: \( \boldsymbol{15.625 \, \text{cm/s}^2} \)