Physics with Calculus: Kinematics and Motion Practice Guidance

Q1. For each set of graphs, fill in the missing graphs and motion descriptions.

Background

Topic: Kinematics Graphs (Position, Velocity, Acceleration)

This question tests your understanding of how position ($x$), velocity ($v$), and acceleration ($a$) graphs relate to each other, and how to describe motion based on these graphs.

Key Terms and Concepts:

• Position ($x$): The location of an object at a given time.

• Velocity ($v$): The rate of change of position with respect to time; the slope of the $x$ vs. $t$ graph.

• Acceleration ($a$): The rate of change of velocity with respect to time; the slope of the $v$ vs. $t$ graph.

Key Relationships:

• If you have a position vs. time graph, the slope at any point gives the velocity at that time.

• If you have a velocity vs. time graph, the slope at any point gives the acceleration at that time.

• The area under the velocity vs. time graph gives the displacement.

Step-by-Step Guidance

1. Examine the given graph(s) and identify which are provided (position, velocity, or acceleration).

2. Determine the relationship between the graphs: for example, if you have a velocity vs. time graph, its slope gives acceleration, and its area gives displacement.

3. Use the motion description (e.g., "move in the negative direction with positive acceleration, beyond the point at which you change direction") to infer the shape and direction of the missing graphs.

4. Sketch or describe the missing graph(s) by considering how the motion changes over time (e.g., if velocity crosses zero, the object changes direction).

Try solving on your own before revealing the answer!

Final Answer:

The missing graphs and descriptions should show consistent relationships: for example, if velocity is negative but acceleration is positive, the object is slowing down in the negative direction, possibly coming to rest and then moving in the positive direction. The position graph would show a minimum at the turning point. The motion description should match the behavior shown in the graphs.

Q2. For the velocity-versus-time graph (1 s per tick), object at $x = -5$ m at $t = 0$ s:

• a. What is the acceleration of the object at $t = 3$ s?

• b. What is the acceleration of the object at $t = 5$ s?

• c. What is the position of the object at $t = 5$ s? (Answer: 1 m)

• d. What is the displacement of the object between $t = 1$ s and $t = 8$ s? (Answer: -5 m)

Background

Topic: Kinematics – Interpreting Velocity-Time Graphs

This question tests your ability to extract acceleration and displacement information from a velocity vs. time graph, and to use initial position to find position at a later time.

Key Terms and Formulas:

• Acceleration ($a$): The slope of the velocity vs. time graph at a given point.

• Displacement ($\Delta x$): The area under the velocity vs. time graph between two times.

• Position at time $t$: $x(t) = x_0 + \int_{0}^{t} v(t') dt'$

Step-by-Step Guidance

1. To find acceleration at a specific time, determine the slope of the $v$ vs. $t$ graph at that time. If the graph is linear, the slope is constant; if not, find the tangent at the point.

2. For position at $t = 5$ s, calculate the area under the $v$ vs. $t$ curve from $t = 0$ to $t = 5$ s. Add this displacement to the initial position $x_0 = -5$ m.

3. For displacement between $t = 1$ s and $t = 8$ s, find the area under the $v$ vs. $t$ curve between these times. Pay attention to the sign (above or below the time axis).

4. Set up the necessary integrals or geometric area calculations (rectangles, triangles, etc.) for the areas under the curve.

Try solving on your own before revealing the answer!

Final Answer:

• a. The acceleration at $t = 3$ s is the slope of the $v$ vs. $t$ graph at that point.

• b. The acceleration at $t = 5$ s is the slope at $t = 5$ s.

• c. The position at $t = 5$ s is $1$m, found by adding the area under the curve from$t = 0$to$t = 5$ s to the initial position.

• d. The displacement between $t = 1$ s and $t = 8$ s is $-5$ m, calculated as the area under the curve between those times.

Q3. For the velocity-versus-time graph (1 s per tick), object at $x = -8$ m at $t = 0$ s:

• a. What is the acceleration of the object at $t = 6$ s?

• b. What is the position of the object at $t = 2$ s? (Answer: -16 m)

• c. What is the position of the object at $t = 4$ s? (Answer: -16 m)

• d. What is the total displacement between $t = 2$ s and $t = 8$ s? (Answer: 9 m)

Background

Topic: Kinematics – Velocity-Time Graphs and Displacement

This question tests your ability to interpret velocity-time graphs to find acceleration, position, and displacement, using initial conditions.

Key Terms and Formulas:

• Acceleration ($a$): Slope of the $v$ vs. $t$ graph at a given time.

• Displacement ($\Delta x$): Area under the $v$ vs. $t$ graph between two times.

• Position at time $t$: $x(t) = x_0 + \int_{0}^{t} v(t') dt'$

Step-by-Step Guidance

1. Find the slope of the $v$ vs. $t$ graph at $t = 6$ s to determine acceleration.

2. To find position at $t = 2$ s, calculate the area under the $v$ vs. $t$ curve from $t = 0$ to $t = 2$ s, and add to $x_0 = -8$ m.

3. Repeat for $t = 4$ s: area under the curve from $t = 0$ to $t = 4$ s, plus $x_0$.

4. For total displacement between $t = 2$ s and $t = 8$ s, calculate the area under the curve between those times.

Try solving on your own before revealing the answer!

Final Answer:

• a. The acceleration at $t = 6$ s is the slope of the $v$ vs. $t$ graph at that point.

• b. The position at $t = 2$ s is $-16$ m.

• c. The position at $t = 4$ s is $-16$ m.

• d. The total displacement between $t = 2$ s and $t = 8$ s is $9$ m.

Q4. A car driving in a straight line slams on the brakes and skids for $0.750$ s, slowing down at a uniform rate of $9.00$ m/s$^2$. The skid marks are $15.0$ m long.

• (a) What is the speed of the car at the moment the brakes are first applied? (Answer: 23.4 m/s)

• (b) What is the speed of the car at the end of the skid? (Answer: 16.6 m/s)

Background

Topic: Kinematics – Uniform Acceleration

This problem tests your ability to use kinematic equations to relate acceleration, time, distance, and velocity for an object slowing down at a constant rate.

Key Terms and Formulas:

• Initial velocity ($v_0$): The speed at which the car begins to skid.

• Final velocity ($v$): The speed at the end of the skid.

• Acceleration ($a$): The rate of change of velocity (negative for slowing down).

• Displacement ($\Delta x$): The distance covered during the skid.

Key Equations:

• $v = v_0 + a t$

• $\Delta x = v_0 t + \frac{1}{2} a t^2$

• $v^2 = v_0^2 + 2 a \Delta x$

Step-by-Step Guidance

1. List the known values: $a = -9.00$ m/s$^2$, $t = 0.750$ s, $\Delta x = 15.0$ m.

2. Use the kinematic equations to relate $v_0$, $v$, $a$, $t$, and $\Delta x$.

3. Set up the equations for $\Delta x$ and $v$ in terms of $v_0$ and solve for $v_0$ first.

4. Once $v_0$ is found, use $v = v_0 + a t$ to find the final speed at the end of the skid.

Try solving on your own before revealing the answer!

Final Answer:

• (a) The speed at the moment the brakes are first applied is $23.4$ m/s.

• (b) The speed at the end of the skid is $16.6$ m/s.

These are found by solving the kinematic equations for $v_0$ and $v$ using the given values.

Q5. A driver has a reaction time of $0.50$ s, and the maximum deceleration of her car is $6.0$ m/s$^2$. She is driving at $20$m/s, when she suddenly sees an obstacle $50$ m in front of her. How much distance does she cover from the instant she sees the obstacle to the instant her car slows to a stop? Does she run over the obstacle? (Answer: 43 m)

Background

Topic: Kinematics – Stopping Distance and Reaction Time

This problem tests your ability to calculate total stopping distance, including both the distance traveled during reaction time and the distance required to stop under constant deceleration.

Key Terms and Formulas:

• Reaction distance: Distance traveled during the driver's reaction time.

• Braking distance: Distance required to stop once brakes are applied.

• Deceleration ($a$): Negative acceleration (slowing down).

Key Equations:

• Reaction distance: $d_{\text{reaction}} = v_0 t_{\text{reaction}}$

• Braking distance: $v^2 = v_0^2 + 2 a d_{\text{brake}}$ (solve for $d_{\text{brake}}$)

Step-by-Step Guidance

1. Calculate the distance the car travels during the reaction time using $d_{\text{reaction}} = v_0 t_{\text{reaction}}$.

2. Set up the kinematic equation for the braking distance, with $v = 0$ (car comes to rest), $v_0 = 20$ m/s, and $a = -6.0$ m/s$^2$.

3. Solve for $d_{\text{brake}}$ using $0 = v_0^2 + 2 a d_{\text{brake}}$.

4. Add the reaction distance and braking distance to get the total stopping distance.

Try solving on your own before revealing the answer!

Final Answer:

The total distance covered is $43$m, which is less than the $50$ m to the obstacle, so the driver does not run over the obstacle.

Q6. A train accelerates from rest at $1.10$ m/s$^2$ for $20.0$ s, then proceeds at constant speed for $1100.0$ m before slowing down at $2.20$ m/s$^2$ until it stops at the station.

• a) What is the distance between the stations? (Answer: 1430 m)

• b) How much time does it take the train to go between the stations? (Answer: 80 s)

Background

Topic: Kinematics – Piecewise Motion (Acceleration, Constant Speed, Deceleration)

This problem tests your ability to analyze motion in segments: acceleration, constant velocity, and deceleration, and to sum distances and times for each segment.

Key Terms and Formulas:

• Acceleration phase: $v = v_0 + a t$, $x = x_0 + v_0 t + \frac{1}{2} a t^2$

• Constant speed phase: $x = v t$

• Deceleration phase: $v^2 = v_0^2 + 2 a x$

Step-by-Step Guidance

1. Calculate the final speed after the acceleration phase using $v = v_0 + a t$.

2. Find the distance covered during acceleration using $x = x_0 + v_0 t + \frac{1}{2} a t^2$.

3. The train then travels $1100.0$ m at constant speed; calculate the time for this segment using $t = x / v$.

4. For the deceleration phase, use $v^2 = v_0^2 + 2 a x$ (with $v = 0$) to find the stopping distance, and then the time to stop.

5. Add up the distances and times for all three segments to get the total distance and total time.

Try solving on your own before revealing the answer!

Final Answer:

• a) The distance between the stations is $1430$ m.

• b) The total time taken is $80$ s.

Q7. A ball is tossed straight up into the air. If exactly $1.2$ s later, it is falling downward at a speed of $0.41$ m/s, with what speed was it tossed upward? (Answer: 11.4 m/s)

Background

Topic: Kinematics – Vertical Motion under Gravity

This problem tests your ability to use kinematic equations for vertical motion, considering the direction of velocity and the effect of gravity.

Key Terms and Formulas:

• Initial velocity ($v_0$): The speed at which the ball is thrown upward.

• Final velocity ($v$): The velocity after $1.2$ s (downward, so negative if up is positive).

• Acceleration due to gravity ($g$): $9.8$ m/s$^2$ downward.

Key Equation:

• $v = v_0 - g t$ (if up is positive)

Step-by-Step Guidance

1. Assign the correct sign to the final velocity (downward is negative if up is positive).

2. Write the kinematic equation relating initial velocity, final velocity, acceleration, and time.

3. Plug in the known values for $v$, $g$, and $t$.

4. Rearrange the equation to solve for $v_0$.

Try solving on your own before revealing the answer!

Final Answer:

The ball was tossed upward with a speed of $11.4$ m/s.

Q8. A ball is tossed straight downward from an initial height of $15.00$ m and strikes the ground at a speed of $30.0$ m/s. Calculate the speed at which it was thrown. (Answer: 24.6 m/s)

Background

Topic: Kinematics – Vertical Motion with Initial Velocity

This problem tests your ability to use kinematic equations to relate initial and final velocities, displacement, and acceleration due to gravity.

Key Terms and Formulas:

• Initial velocity ($v_0$): The speed at which the ball is thrown downward.

• Final velocity ($v$): The speed when the ball hits the ground.

• Displacement ($\Delta y$): The vertical distance fallen ($-15.00$ m if down is negative).

• Acceleration due to gravity ($g$): $9.8$ m/s$^2$ downward.

Key Equation:

• $v^2 = v_0^2 + 2 g \Delta y$

Step-by-Step Guidance

1. Assign the correct signs to all quantities (choose a positive direction, usually downward).

2. Write the kinematic equation relating $v$, $v_0$, $g$, and $\Delta y$.

3. Plug in the known values for $v$, $g$, and $\Delta y$.

4. Rearrange the equation to solve for $v_0$.

Try solving on your own before revealing the answer!

Final Answer:

The ball was thrown downward at a speed of $24.6$ m/s.

Q9. A stone is dropped from rest from the top of a tall tower and takes $4.0$ s to fall to the ground. How tall is the tower? (Answer: 78 m)

Background

Topic: Kinematics – Free Fall

This problem tests your ability to use kinematic equations for objects dropped from rest under gravity.

Key Terms and Formulas:

• Initial velocity ($v_0$): $0$ (dropped from rest).

• Time ($t$): $4.0$ s.

• Acceleration due to gravity ($g$): $9.8$ m/s$^2$ downward.

• Displacement ($h$): The height of the tower.

Key Equation:

• $h = v_0 t + \frac{1}{2} g t^2$

Step-by-Step Guidance

1. Since the stone is dropped from rest, $v_0 = 0$.

2. Write the kinematic equation for displacement under constant acceleration.

3. Plug in the values for $g$ and $t$.

4. Solve for $h$ (height of the tower).

Try solving on your own before revealing the answer!

Final Answer:

The tower is $78$ m tall.

Q10. A ball is tossed straight downward from an initial height of $15.00$ m and strikes the ground at a speed of $30.0$ m/s. Calculate the speed at which it was thrown. (Answer: 24.6 m/s)

Background

Topic: Kinematics – Vertical Motion with Initial Velocity

This is a repeat of Q8, testing your ability to use kinematic equations for vertical motion with initial velocity.

Key Terms and Formulas:

• See Q8 for terms and equations.

Step-by-Step Guidance

1. Follow the same steps as in Q8: assign signs, write the kinematic equation, plug in values, and solve for $v_0$.

Try solving on your own before revealing the answer!

Final Answer:

The ball was thrown downward at a speed of $24.6$ m/s.

Q11. An elevator descends at a constant speed of $5.00$ m/s for $3.00$ s and then enters free-fall for $1.50$ s. Calculate:

• (a) The total distance descended during the entire $4.50$ s. (Answer: 33.5 m)

• (b) The speed of the elevator at the end of the $4.50$ s interval. (Answer: 19.7 m/s)

Background

Topic: Kinematics – Piecewise Motion (Constant Speed and Free Fall)

This problem tests your ability to analyze motion in segments: constant velocity and then free fall, and to sum distances and velocities for each segment.

Key Terms and Formulas:

• Constant speed phase: $d_1 = v t$

• Free-fall phase: $d_2 = v_0 t + \frac{1}{2} g t^2$

• Final speed after free fall: $v = v_0 + g t$

Step-by-Step Guidance

1. Calculate the distance descended during the constant speed phase: $d_1 = 5.00$ m/s $\times$ $3.00$ s.

2. At the start of free fall, the elevator's speed is $5.00$ m/s downward. Use this as $v_0$ for the free-fall phase.

3. Calculate the distance descended during free fall: $d_2 = v_0 t + \frac{1}{2} g t^2$ with $t = 1.50$ s.

4. Calculate the final speed at the end of free fall: $v = v_0 + g t$.

5. Add the distances from both phases to get the total distance descended.

Try solving on your own before revealing the answer!

Final Answer:

• (a) The total distance descended is $33.5$ m.

• (b) The speed at the end of $4.50$ s is $19.7$ m/s.

Q12. At $t = 0$ s, a puck of clay is sliding over the ice at $8.00$ m/s, slowing uniformly at $0.500$ m/s$^2$. At $t = 25.0$ s, the puck hits the wall and sticks. What total distance did the puck slide between $t = 0$ s and $t = 25.0$ s? (Answer: 43.75 m)

Background

Topic: Kinematics – Uniform Deceleration

This problem tests your ability to use kinematic equations to find displacement when an object slows down at a constant rate.

Key Terms and Formulas:

• Initial velocity ($v_0$): $8.00$ m/s.

• Acceleration ($a$): $-0.500$ m/s$^2$ (slowing down).

• Time ($t$): $25.0$ s.

• Displacement ($x$): $x = v_0 t + \frac{1}{2} a t^2$

Step-by-Step Guidance

1. Write the kinematic equation for displacement with constant acceleration.

2. Plug in the values for $v_0$, $a$, and $t$.

3. Calculate each term separately before adding them together.

Try solving on your own before revealing the answer!

Final Answer:

The total distance the puck slid is $43.75$ m.