Question

30 as your friend skateboards down a hill, you measure their velocity at 1.0 - second intervals. you record the following data: (0.0 s, 0.0 m/s), (1.0 s, 2.3 m/s), (2.0 s, 4.6 m/s). determine your friend’s acceleration.
31 you stop recording velocity data at t = 4.0 s, but you notice a short time later that your friend is going a very fast 13.8 m/s. assuming your friend keeps the same acceleration, how much time has passed since you stopped recording data?
32 theme patterns when analyzing graphs, the area under a velocity vs. time graph is the displacement. what do you think the area under an acceleration vs. time graph would represent? what would you suppose about the area under a jerk vs. time graph?
33 sep develop a model an equation describing the velocity of an object is ( v = (-2.0 \text{m/s}^2)t + 2.0 \text{m/s} ). sketch by hand or compute the velocity vs. time graph for that motion from 0.0 second to 2.0 seconds.

Explanation:

Question 30

Step1: Recall acceleration formula

Acceleration \( a \) is the change in velocity \( \Delta v \) over the change in time \( \Delta t \), so \( a=\frac{\Delta v}{\Delta t} \).

Step2: Identify velocity and time changes

Initial velocity \( v_0 = 0.0\ m/s \) at \( t_0 = 0.0\ s \), final velocity \( v = 4.6\ m/s \) at \( t = 2.0\ s \). So \( \Delta v=4.6 - 0.0 = 4.6\ m/s \), \( \Delta t=2.0 - 0.0 = 2.0\ s \).

Step3: Calculate acceleration

Substitute into the formula: \( a=\frac{4.6}{2.0}=2.3\ m/s^2 \).

Step1: Recall acceleration from Q30

From Q30, \( a = 2.3\ m/s^2 \). At \( t = 4.0\ s \), velocity \( v_1 \) can be found by \( v = v_0+at \) (here \( v_0 = 0 \)), so \( v_1=2.3\times4.0 = 9.2\ m/s \).

Step2: Use velocity formula for time

We know final velocity \( v_2 = 13.8\ m/s \), initial velocity (at \( t = 4.0\ s \)) \( v_1 = 9.2\ m/s \), acceleration \( a = 2.3\ m/s^2 \). Using \( v_2=v_1+at' \), solve for \( t' \): \( t'=\frac{v_2 - v_1}{a}=\frac{13.8 - 9.2}{2.3}=\frac{4.6}{2.3}=2.0\ s \).

The area under a velocity - time graph is displacement (\( \Delta x=\int v\mathrm{d}t \)). By analogy, acceleration \( a=\frac{\Delta v}{\Delta t} \), so the area under an acceleration - time graph (\( \int a\mathrm{d}t \)) should represent the change in velocity (\( \Delta v \)). Jerk \( j=\frac{\Delta a}{\Delta t} \), so the area under a jerk - time graph (\( \int j\mathrm{d}t \)) should represent the change in acceleration (\( \Delta a \)).

Answer:

\( 2.3\ m/s^2 \)

Question 31