Physics with Calculus: Step-by-Step Guidance for Example Quiz 1

Q1. The density of a material is 7.50 g/cm³. Express this in kg/m³.

Background

Topic: Unit Conversion (Density)

This question tests your ability to convert units for density from grams per cubic centimeter to kilograms per cubic meter, a common skill in introductory physics.

Key Terms and Formulas:

• Density ($\rho$): mass per unit volume.

• Unit conversions: $1\ \text{g} = 10^{-3}\ \text{kg}$, $1\ \text{cm}^3 = 10^{-6}\ \text{m}^3$

Step-by-Step Guidance

1. Start with the given density: $7.50\ \text{g/cm}^3$.

2. Convert grams to kilograms: $7.50\ \text{g} = 7.50 \times 10^{-3}\ \text{kg}$.

3. Convert cubic centimeters to cubic meters: $1\ \text{cm}^3 = 10^{-6}\ \text{m}^3$.

4. Set up the conversion: $7.50\ \text{g/cm}^3 = \frac{7.50 \times 10^{-3}\ \text{kg}}{10^{-6}\ \text{m}^3}$.

Try solving on your own before revealing the answer!

Final Answer: $7.50 \times 10^3$ kg/m³

By dividing $7.50 \times 10^{-3}$ kg by $10^{-6}$ m³, you get $7.50 \times 10^3$ kg/m³.

This conversion is essential for working with SI units in physics.

Q2. Find the magnitude of the acceleration (in m/s²) from the velocity-time graph.

Background

Topic: Kinematics (Acceleration from Velocity-Time Graph)

This question tests your ability to determine acceleration from a velocity vs. time graph, a fundamental concept in motion analysis.

Key Terms and Formulas:

• Acceleration ($a$): rate of change of velocity with respect to time.

• Formula: $a = \frac{\Delta v}{\Delta t}$

Step-by-Step Guidance

1. Identify two points on the graph where the line passes through grid intersections (e.g., $(t_1, v_1)$ and $(t_2, v_2)$).

2. Calculate the change in velocity: $\Delta v = v_2 - v_1$.

3. Calculate the change in time: $\Delta t = t_2 - t_1$.

4. Set up the formula for acceleration: $a = \frac{\Delta v}{\Delta t}$.

Try solving on your own before revealing the answer!

Final Answer: 2.5 m/s²

Using the points (20 m/s, 4 s) and (10 m/s, 8 s): $a = \frac{10 - 20}{8 - 4} = \frac{-10}{4} = -2.5$ m/s². The magnitude is 2.5 m/s².

The negative sign indicates the car is slowing down, but the question asks for magnitude.

Q3. Is the car speeding up or slowing down?

Background

Topic: Kinematics (Velocity and Acceleration)

This question tests your understanding of how velocity changes over time and what it means for an object to speed up or slow down.

Key Terms:

• Speeding up: velocity increases in magnitude.

• Slowing down: velocity decreases in magnitude.

Step-by-Step Guidance

1. Observe the slope of the velocity-time graph: is velocity increasing or decreasing as time progresses?

2. If the velocity decreases as time increases, the object is slowing down.

Try solving on your own before revealing the answer!

Final Answer: Slowing down

The velocity decreases as time increases, so the car is slowing down.

Q4. What is the velocity at $t = 1.0$ s for $x(t) = 3.0t^3 - 5.0t^2 + 2.0t$?

Background

Topic: Calculus in Kinematics (Velocity from Position Function)

This question tests your ability to use calculus to find velocity from a position function.

Key Terms and Formulas:

• Position function: $x(t)$

• Velocity: $v(t) = \frac{dx}{dt}$

Step-by-Step Guidance

1. Write the position function: $x(t) = 3.0t^3 - 5.0t^2 + 2.0t$.

2. Take the derivative with respect to $t$ to find velocity: $v(t) = \frac{dx}{dt}$.

3. Compute the derivative: $v(t) = 9.0t^2 - 10.0t + 2.0$.

4. Substitute $t = 1.0$ s into the velocity function to find $v(1.0)$.

Try solving on your own before revealing the answer!

Final Answer: 1 m/s

Plugging in $t = 1.0$ s: $v(1.0) = 9.0(1.0)^2 - 10.0(1.0) + 2.0 = 1$ m/s.

This matches the tangent to the curve at $t = 1$ s.

Q5. What is the instantaneous acceleration at $t = 2.0$ s for $x(t) = 3.0t^3 - 5.0t^2 + 2.0t$?

Background

Topic: Calculus in Kinematics (Acceleration from Position Function)

This question tests your ability to use calculus to find acceleration from a position function.

Key Terms and Formulas:

• Acceleration: $a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

• Velocity function: $v(t) = 9.0t^2 - 10.0t + 2.0$

Step-by-Step Guidance

1. Take the derivative of the velocity function to find acceleration: $a(t) = \frac{dv}{dt}$.

2. Compute the derivative: $a(t) = 18.0t - 10.0$.

3. Substitute $t = 2.0$ s into the acceleration function to find $a(2.0)$.

Try solving on your own before revealing the answer!

Final Answer: 26 m/s²

Plugging in $t = 2.0$ s: $a(2.0) = 18.0(2.0) - 10.0 = 26$ m/s².

This is the instantaneous acceleration at $t = 2.0$ s.

Q9. Three displacement vectors are drawn along edges/diagonals of a rectangle as shown. Which relation is correct?

Background

Topic: Vector Addition and Subtraction

This question tests your understanding of vector addition and subtraction using geometric representations.

Key Terms and Formulas:

• Vector addition: tip-to-tail method

• Vector subtraction: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$

Step-by-Step Guidance

1. Examine the diagram and identify the starting and ending points of each vector.

2. Use the tip-to-tail method to see how combining vectors $\vec{A}$ and $\vec{B}$ relates to $\vec{C}$.

3. Try rearranging the vectors algebraically to match the geometric path shown in the diagram.

Try solving on your own before revealing the answer!

Final Answer: $\vec{B} - \vec{A} = \vec{C}$

Following $\vec{B}$ and then $-\vec{A}$ (leftward) gives the same displacement as $\vec{C}$, so $\vec{B} - \vec{A} = \vec{C}$.