Question

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

Explanation:

Assuming the question is to find the time taken or acceleration (since it's a uniformly accelerated motion, as speed increases steadily). Let's assume we need to find the time taken first, using the formula for average speed.

Step1: Recall average speed formula

For uniformly accelerated motion (starting from rest, \( s_i = 0 \)), average speed \( v_{avg}=\frac{s_i + s_f}{2} \), and distance \( d = v_{avg} \times t \).
Given \( d = 80 \, \text{cm} \), \( s_i = 0 \, \text{cm/s} \), \( s_f = 50 \, \text{cm/s} \).
First, calculate average speed: \( v_{avg}=\frac{0 + 50}{2}=25 \, \text{cm/s} \)

Step2: Solve for time \( t \)

Using \( d = v_{avg} \times t \), rearrange to \( t=\frac{d}{v_{avg}} \)
Substitute \( d = 80 \, \text{cm} \), \( v_{avg}=25 \, \text{cm/s} \): \( t=\frac{80}{25}=3.2 \, \text{s} \)
(If we want acceleration, use \( v = u + at \), \( u = 0 \), \( v = 50 \), \( t = 3.2 \), so \( a=\frac{50 - 0}{3.2}=15.625 \, \text{cm/s}^2 \))

Answer:

If finding time, the time taken is \( 3.2 \, \text{seconds} \); if finding acceleration, acceleration is \( 15.625 \, \text{cm/s}^2 \) (depending on the exact question, but assuming time first, answer is \( 3.2 \, \text{s} \))