Question

1. a physics student skis down a slope, with a constant acceleration of 2.0 m/s² for 15 seconds.

a. make a well - labeled diagram of the situation.
b. make a well - labeled graphical representation of the situation.
c. list given quantities and quantities to find as you determine the length of the slope.

Explanation:

Part (a) - Diagram

Step1: Draw Slope

Draw a slanted line (slope) with an angle to the horizontal. Label the starting point as the origin of motion.

Step2: Add Student and Vectors

Draw a skier at the top of the slope. Add an acceleration vector (\( \vec{a} \)) parallel to the slope, pointing down. Label acceleration magnitude \( a = 2.0\ \text{m/s}^2 \), time \( t = 15\ \text{s} \), and the slope length \( d \) (to find).

Part (b) - Graph (Velocity - Time)

Step1: Axes Setup

On the given grid, label the y - axis as "Velocity (\( \text{m/s} \))" and x - axis as "Time (\( \text{s} \))". The initial velocity \( u = 0\ \text{m/s} \) (starts from rest), so the graph starts at (0, 0).

Step2: Plot Line

Since \( a=\frac{\Delta v}{\Delta t} \), \( v = u+at \). With \( u = 0 \), \( v=2t \). At \( t = 15\ \text{s} \), \( v=2\times15 = 30\ \text{m/s} \). Draw a straight line from (0, 0) to (15, 30). The area under this line (a triangle) represents displacement (slope length).

Part (c) - Quantities and Calculation

Step1: List Given

• Initial velocity, \( u = 0\ \text{m/s} \) (starts from rest),
• Acceleration, \( a = 2.0\ \text{m/s}^2 \),
• Time, \( t = 15\ \text{s} \).
• Quantity to find: Length of slope (\( d \), displacement).

Step2: Use Kinematic Formula

The kinematic equation for displacement with constant acceleration is \( d=ut+\frac{1}{2}at^{2} \).

Step3: Substitute Values

Substitute \( u = 0 \), \( a = 2.0\ \text{m/s}^2 \), \( t = 15\ \text{s} \) into the formula:
\( d=0\times15+\frac{1}{2}\times2.0\times(15)^{2} \)
\( d = 0+\frac{1}{2}\times2\times225 \)
\( d=225\ \text{m} \)

Answer:

s:
a. Diagram: Slanted line with skier, acceleration vector, and labels as described.
b. Graph: Velocity - time graph with line from (0,0) to (15,30), area under line is displacement.
c. Given: \( u = 0\ \text{m/s} \), \( a = 2.0\ \text{m/s}^2 \), \( t = 15\ \text{s} \); To find: \( d \). Length of slope: \( d = 225\ \text{m} \)