Question

1. a ball rolls down a hill and its velocity over time is recorded in the table below. determine the acceleration of the ball.

velocity	0 m/s	5 m/s	10 m/s	15 m/s
time	0 seconds	1 seconds	2 seconds	3 seconds

a. 0 m/s²
b. 1 m/s²
c. 3 m/s²
d. 5 m/s²
e. 15 m/s²
use the graph below to answer questions 22 - 24.
motion of a cat

1. according to the graph above, during which interval is the cat at rest?

a. 0.0 - 5.0 s b. 5.0 - 10.0 s c. 10.0 - 15.0 s d. 15.0 - 20.0 s

1. according to the graph above, the cat has the fastest speed during which interval?

a. 0.0 - 5.0 s b. 5.0 - 10.0 s c. 10.0 - 15.0 s d. 15.0 - 20.0 s

1. according to the graph above, how far does the cat end up from its original starting location?

a. same location b. 3 m c. 1.5 m d. 0.5 m

1. a baseball player throws a ball straight upwards with an initial velocity of “v₀” m/s. the ball experiences an acceleration due to gravity equal to “g” m/s². determine an expression for the maximum height achieved by the ball in meters.

a. $\frac{v_{0}}{2g}$ b. $2gv_{0}$ c. $sqrt{2gv_{0}}$ d. $\frac{-v_{0}^{2}}{2g}$

Explanation:

Step1: Recall acceleration formula

Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v$ is change in velocity and $\Delta t$ is change in time.

Step2: Select two - point data

Let's take the first and the last data points. Initial velocity $v_1 = 0$ m/s at $t_1=0$ s, final velocity $v_2 = 15$ m/s at $t_2 = 3$ s.

Step3: Calculate $\Delta v$ and $\Delta t$

$\Delta v=v_2 - v_1=15 - 0=15$ m/s, $\Delta t=t_2 - t_1=3 - 0 = 3$ s.

Step4: Calculate acceleration

$a=\frac{\Delta v}{\Delta t}=\frac{15}{3}=5$ m/s².

Step1: Understand rest in displacement - time graph

In a displacement - time graph, an object is at rest when the displacement does not change with time (slope of the graph is 0).

Step2: Analyze intervals

In the interval 5.0 - 10.0 s, the displacement of the cat remains constant at 3.0 m. The slope of the graph in this interval is 0.

Step1: Recall speed and slope of displacement - time graph

Speed is the magnitude of velocity, and in a displacement - time graph, the slope of the graph represents velocity. Steeper the slope, higher the speed.

Step2: Analyze intervals

The slope is steepest in the interval 10.0 - 15.0 s.

Step1: Determine initial and final displacement

Initial displacement at $t = 0$ s is 2.0 m, and final displacement at $t=20$ s is 0.5 m.

Step2: Calculate the difference

The cat ends up $2.0 - 0.5=1.5$ m from its original starting location.

Answer:

d. 5 m/s²