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.
sᵢ =
unit
a =
unit
s_f =
unit
d =
unit
t =
unit
how long did the marble take to reach the bottom of the ramp?
unit

Explanation:

Step1: Identify knowns

The marble starts from rest, so initial speed \( s_i = 0 \, \text{cm/s} \). Final speed \( s_f = 50 \, \text{cm/s} \). Distance \( d = 80 \, \text{cm} \). For uniformly accelerated motion (steady speed increase), we can use the average speed formula. Average speed \( v_{avg}=\frac{s_i + s_f}{2} \).

Step2: Calculate average speed

\( v_{avg}=\frac{0 + 50}{2}=25 \, \text{cm/s} \)

Step3: Use distance formula \( d = v_{avg} \times t \) to find time \( t \)

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

Filling the unknowns:

• \( s_i = 0 \, \text{cm/s} \) (initial speed, starts from rest)
• \( s_f = 50 \, \text{cm/s} \) (final speed)
• \( d = 80 \, \text{cm} \) (distance of the ramp)
• \( t \): We found \( t = 3.2 \, \text{s} \) (time to reach bottom)

Answer:

For the time taken: \( 3.2 \, \text{seconds} \) (or \( 3.2 \, \text{s} \))

Filling the boxes:

• \( s_i = \boldsymbol{0} \), unit: \(\boldsymbol{\text{cm/s}}\)
• \( s_f = \boldsymbol{50} \), unit: \(\boldsymbol{\text{cm/s}}\)
• \( d = \boldsymbol{80} \), unit: \(\boldsymbol{\text{cm}}\)
• \( t = \boldsymbol{3.2} \), unit: \(\boldsymbol{\text{s}}\) (for the time to reach bottom, the answer is \( 3.2 \) with unit \(\text{s}\))