Question

34 theme patterns the equations of motion can only be used when acceleration treated as constant. for comparison, sketch by hand or computer and label example velocity vs. time graphs for the two cases—constant accelerated motion and accelerated motion that is not constant.
35 sep use mathematics some steps in determining the area of the triangle are missing. use the velocity equation of motion to show that ( v - v_i ) is equal to ( at ). then, show that the area of the triangle can be written as ( \frac{1}{2}at^2 ).
36 sep construct an explanation go online to find a graph for accelerated motion that is not constant. use the graph to explain why the displacement equation of motion is not appropriate for analyzing accelerated motion that is not constant.
37 sep obtain and evaluate information often, a third equation of motion is used when solving constant acceleration problems: ( v^2 = v_i^2 + 2adelta d ). go online to find how the equation is derived. does it come directly from analyzing the motion graphs, or is it derived from the other two equations of motion? when would it be useful?
38 at ( t = 0 )s, a bunny is hopping along at 2.0 m/s. it continues at that speed until at ( t = 20 )s, when the bunny begins to slow down with a constant acceleration. it comes to a complete stop at ( t = 25 )s. what is the bunny’s distance traveled at 25 seconds? what is the bunny’s acceleration when slowing down?

Explanation:

Problem 35 Solution:

Step 1: Recall the velocity equation of motion

The velocity equation for an object with constant acceleration \( a \) is given by:
\( v = v_i + at \)
where \( v \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

Step 2: Derive \( v - v_i = at \)

Subtract \( v_i \) from both sides of the velocity equation:
\( v - v_i = v_i + at - v_i \)
Simplifying the right-hand side:
\( v - v_i = at \)

Step 3: Recall the formula for the area of a triangle

The area of a triangle is given by:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

Step 4: Identify base and height for the velocity-time triangle

For a velocity-time graph of constant acceleration, the triangle (representing displacement) has:

• Base: time \( t \)
• Height: \( v - v_i \) (which we just showed is \( at \))

Step 5: Substitute height into the area formula

Substitute \( \text{height} = at \) and \( \text{base} = t \) into the area formula:
\( \text{Area} = \frac{1}{2} \times t \times (at) \)
Simplify the expression:
\( \text{Area} = \frac{1}{2}at^2 \)

Step 1: Analyze the bunny’s motion in two phases

• Phase 1 (Constant Speed): From \( t = 0 \) to \( t = 20 \, \text{s} \), the bunny moves at a constant speed \( v = 2.0 \, \text{m/s} \).
• Phase 2 (Constant Deceleration): From \( t = 20 \, \text{s} \) to \( t = 25 \, \text{s} \), the bunny slows down with constant acceleration (deceleration) until it stops.

Step 2: Calculate distance in Phase 1

Use the formula for distance with constant speed:
\( d_1 = v \times t_1 \)
where \( t_1 = 20 \, \text{s} - 0 \, \text{s} = 20 \, \text{s} \), and \( v = 2.0 \, \text{m/s} \).

Substitute values:
\( d_1 = 2.0 \, \text{m/s} \times 20 \, \text{s} = 40 \, \text{m} \)

Step 3: Calculate distance in Phase 2

For Phase 2, the bunny decelerates from \( v_i = 2.0 \, \text{m/s} \) to \( v_f = 0 \, \text{m/s} \) over \( t_2 = 25 \, \text{s} - 20 \, \text{s} = 5 \, \text{s} \).

Use the average velocity formula for constant acceleration:
\( d_2 = \frac{v_i + v_f}{2} \times t_2 \)

Substitute values:
\( d_2 = \frac{2.0 \, \text{m/s} + 0 \, \text{m/s}}{2} \times 5 \, \text{s} = 1.0 \, \text{m/s} \times 5 \, \text{s} = 5 \, \text{m} \)

Step 4: Total distance traveled

Add the distances from both phases:
\( d_{\text{total}} = d_1 + d_2 = 40 \, \text{m} + 5 \, \text{m} = 45 \, \text{m} \)

Step 5: Calculate acceleration in Phase 2

Use the velocity equation \( v_f = v_i + a t_2 \). Rearrange for \( a \):
\( a = \frac{v_f - v_i}{t_2} \)

Substitute values (\( v_f = 0 \, \text{m/s} \), \( v_i = 2.0 \, \text{m/s} \), \( t_2 = 5 \, \text{s} \)):
\( a = \frac{0 - 2.0 \, \text{m/s}}{5 \, \text{s}} = -0.4 \, \text{m/s}^2 \)
(The negative sign indicates deceleration.)

Answer:

We derived \( v - v_i = at \) from the velocity equation of motion \( v = v_i + at \). For the triangle’s area, using the base \( t \) and height \( at \) (from \( v - v_i = at \)), we substituted into the triangle area formula \( \frac{1}{2} \times \text{base} \times \text{height} \) to get \( \frac{1}{2}at^2 \).

Problem 38 Solution: