Question

34 theme patterns the equations of motion can only be used when acceleration is 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? 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 (the last one about the bunny):

We need to find the distance traveled by the bunny at \( t = 25\space s \) and its acceleration when slowing down.

Step 1: Analyze the motion phases

The bunny's motion has two phases:

1. Constant speed phase (from \( t = 0\space s \) to \( t = 20\space s \)): The speed \( v_1 = 2.0\space m/s \), and the time duration \( t_1 = 20\space s - 0\space s = 20\space s \).
2. Deceleration phase (from \( t = 20\space s \) to \( t = 25\space s \)): The initial speed for this phase is \( v_{02}=2.0\space m/s \), the final speed \( v_{2}=0\space m/s \) (comes to a stop), and the time duration \( t_2 = 25\space s - 20\space s = 5\space s \).

Step 2: Calculate the distance in the constant - speed phase

The formula for distance when speed is constant is \( d = v\times t \).
For the first phase, \( d_1=v_1\times t_1 \)
Substitute \( v_1 = 2.0\space m/s \) and \( t_1 = 20\space s \) into the formula:
\( d_1=2.0\space m/s\times20\space s = 40\space m \)

Step 3: Calculate the distance in the deceleration phase

We can use the formula for the distance of an object with constant acceleration (deceleration in this case) \( d_2=\frac{v_{02} + v_2}{2}\times t_2 \) (this formula comes from the fact that the average speed \( \bar{v}=\frac{v_{0}+v}{2} \) for constant acceleration motion, and \( d=\bar{v}\times t \)).
Substitute \( v_{02}=2.0\space m/s \), \( v_2 = 0\space m/s \) and \( t_2 = 5\space s \) into the formula:
\( d_2=\frac{2.0\space m/s+ 0\space m/s}{2}\times5\space s=1.0\space m/s\times5\space s = 5\space m \)

Step 4: Calculate the total distance

The total distance \( d = d_1 + d_2 \)
\( d=40\space m + 5\space m=45\space m \)

Step 5: Calculate the acceleration during deceleration

We use the formula \( v = v_0+at \) (rearranged to \( a=\frac{v - v_0}{t} \)).
Here, \( v = v_2 = 0\space m/s \), \( v_0 = v_{02}=2.0\space m/s \), and \( t = t_2 = 5\space s \)
\( a=\frac{0\space m/s - 2.0\space m/s}{5\space s}=\frac{- 2.0\space m/s}{5\space s}=- 0.4\space m/s^{2} \)
The negative sign indicates that it is a deceleration (acceleration opposite to the direction of motion).

Answer:

s:

• The distance traveled by the bunny at \( t = 25\space s \) is \( \boldsymbol{45\space m} \).
• The acceleration of the bunny when slowing down is \( \boldsymbol{- 0.4\space m/s^{2}} \) (or a deceleration of \( 0.4\space m/s^{2} \)).