AP Physics 1: Algebra-Based Practice Quiz: Representing Motion

Correct Answer: C

The provided content explicitly lists motion diagrams, figures, graphs, equations, and narrative descriptions as ways to represent motion. Temperature maps are not mentioned as a representation of motion.

Correct Answer: B

According to the provided content, an object's instantaneous velocity is the rate of change of the object's position, which is equal to the slope of a line tangent to a point on a graph of the object's position as a function of time.

Correct Answer: C

The content states that an object's instantaneous acceleration is the rate of change of the object's velocity, which is equal to the slope of a line tangent to a point on a graph of the object's velocity as a function of time.

Correct Answer: C

The provided text explicitly states that the displacement of an object during a time interval is equal to the area under the curve of a graph of the object's velocity as a function of time.

Correct Answer: C

As stated in the content, the change in velocity of an object during a time interval is equal to the area under the curve of a graph of the acceleration of the object as a function of time.

Correct Answer: A

The equation $v_x = v_{x0} + a_x t$ directly relates final velocity ($v_x$) to initial velocity ($v_{x0}$), acceleration ($a_x$), and time ($t$), which are the known quantities and the desired unknown.

Correct Answer: B

Using the kinematic equation $x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2$. Given $x_0 = 0$, $v_{x0} = 0$ (starts from rest), $a_x = 4 \text{ m/s}^2$, and $t = 3 \text{ s}$. Plugging in the values: $x = 0 + (0)(3) + \frac{1}{2} (4) (3)^2 = \frac{1}{2} (4)(9) = 2 \times 9 = 18 \text{ m}$.

Correct Answer: C

Near the surface of Earth, the vertical acceleration caused by gravity is constant and directed downward, with a value of approximately 10 m/s². This acceleration is constant throughout the flight of the object, including at the very top of its path where its instantaneous velocity is zero.

Correct Answer: B

The slope of a position-time graph represents velocity. A straight line has a constant slope. Therefore, a straight line with a positive slope on a position-time graph indicates a constant positive velocity. Since the velocity is constant, the acceleration is zero.

Correct Answer: D

The slope of the velocity-versus-time graph represents the instantaneous acceleration. If the curve is concave up, its slope is continuously increasing. Therefore, the object's acceleration is increasing.

Correct Answer: B

Use the kinematic equation $v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)$. Here, $v_{x0} = 5 \text{ m/s}$, $v_x = 15 \text{ m/s}$, and $(x - x_0) = 50 \text{ m}$. Rearranging for $a_x$: $a_x = \frac{v_x^2 - v_{x0}^2}{2(x - x_0)} = \frac{(15)^2 - (5)^2}{2(50)} = \frac{225 - 25}{100} = \frac{200}{100} = 2 \text{ m/s}^2$.

Correct Answer: B

We can use the kinematic equation $v_x = v_{x0} + a_x t$. In this case, the object is dropped from rest, so $v_{x0} = 0$. The acceleration is due to gravity, $a_x = g \approx 10 \text{ m/s}^2$. The time is $t = 2 \text{ s}$. Therefore, $v_x = 0 + (10)(2) = 20 \text{ m/s}$.

Correct Answer: B

"Moving in the negative direction" means the velocity is negative. "Slowing down" means that the acceleration is in the opposite direction of the velocity. The opposite of a negative velocity is a positive acceleration.

Correct Answer: B

The displacement is the area under the velocity-versus-time graph. The described graph forms a triangle with a base of 4 s and a height of 8 m/s. The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. Therefore, the displacement is $\frac{1}{2} \times (4 \text{ s}) \times (8 \text{ m/s}) = 16 \text{ m}$.

Correct Answer: B

The equation $x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2$ allows for the calculation of displacement (by rearranging to $x - x_0 = v_{x0} t + \frac{1}{2} a_x t^2$) using initial velocity, acceleration, and time, without needing the final velocity. The other equations either solve for final velocity or require it as an input.

Correct Answer: A

The acceleration is the slope of the velocity-time graph. A horizontal line has a slope of zero, so the acceleration is 0 m/s². The displacement is the area under the curve. The area is a rectangle with height 5 m/s and width 4 s. Area = height × width = (5 m/s) × (4 s) = 20 m.