Q1. A cheetah, the world’s fastest land mammal, has a maximum acceleration of 8.9 m/s2 and a top speed of 110 km/h (about 70 mph). Starting from rest, what distance can it run in 6.00 s?

Background

Topic: 1-D Kinematics with Constant Acceleration

This question tests your ability to apply the equations of motion for constant acceleration to determine the distance traveled by an object starting from rest.

Key Terms and Formulas

• Initial velocity (): The velocity at the start (here, 0 m/s).

• Acceleration (): The rate of change of velocity (here, 8.9 m/s2).

• Time (): The duration of motion (here, 6.00 s).

• Displacement (): The distance traveled.

Key formula:

Step-by-Step Guidance

1. Identify the known values: m/s, m/s2, s, and (assuming the cheetah starts at the origin).

2. Write the kinematic equation for displacement with constant acceleration:

3. Since the cheetah starts from rest, , so the equation simplifies to:

4. Plug in the values for and (but do not calculate the final value yet):

Try solving on your own before revealing the answer!

Q2. A rocket sits initially at rest on the surface of the earth, until it begins burning its engines. While burning its engines, it accelerates upward at 30.0 m/s2 for 30.0 s. It then runs out of fuel and continues ascending to its maximum height in free fall. Ignoring air resistance, (i) what is the rocket’s maximum altitude, and (ii) how much time passes from liftoff to reaching its maximum altitude?

Background

Topic: 1-D Kinematics with Multiple Acceleration Phases (Piecewise Motion)

This problem involves two stages: (1) powered ascent with constant acceleration, and (2) free-fall ascent after fuel runs out. You must analyze each phase separately and combine results.

Key Terms and Formulas

• Initial velocity (), acceleration (), time (), displacement ()

• For constant acceleration:

• Final velocity after powered ascent:

• For free-fall: (where m/s2)

• Maximum height: occurs when during free-fall phase

Step-by-Step Guidance

1. First, calculate the velocity at the end of the powered ascent using:

   where , m/s2, s.

2. Calculate the displacement during the powered ascent:

3. For the free-fall phase, use the velocity at the end of powered ascent as the initial velocity upward. The rocket will continue to rise until its velocity becomes zero.

   Set and use:

   where .

4. Solve for the time it takes to reach maximum height after fuel runs out.

5. Calculate the additional displacement during free-fall ascent:

6. Add and to get the total maximum altitude. Add and for total time to maximum altitude.

Try solving on your own before revealing the answer!

Q3. An elevator sits at rest at the top floor of a skyscraper, when the cable suddenly snaps, and it starts plummeting downward. It falls freely for 1.20 s before a safety mechanism is activated that exerts a friction force exactly balancing the weight of the elevator and stopping its acceleration, so that it descends at a constant speed thereafter. Calculate the total distance the elevator descends during a time interval 4.00 s after the cable snaps.

Background

Topic: Piecewise 1-D Kinematics (Free-Fall and Uniform Motion)

This problem involves two phases: (1) free-fall with constant acceleration, and (2) constant velocity motion after the safety mechanism activates. You must analyze each phase and sum the distances.

Key Terms and Formulas

• Free-fall acceleration: m/s2

• Displacement during free-fall:

• Velocity at end of free-fall:

• Displacement during constant velocity:

• Total time: s; s;

Step-by-Step Guidance

1. Calculate the distance fallen during the first 1.20 s of free-fall:

2. Find the velocity at the end of free-fall (just before the safety mechanism activates):

3. Determine the time spent descending at constant speed: s.

4. Calculate the distance descended at constant speed:

5. Add and to get the total distance descended.

Try solving on your own before revealing the answer!

Q4. A runner jogs for 150 m due west (displacement A) and then turns to a direction 30.0 degrees west of south and jogs for another 90.0 m (displacement B). Finally, she turns to a direction 20.0 degrees south of east and jogs for another 80.0 m (displacement C). Showing all work and units, and putting boxes around your answers, calculate (a) the total distance between her initial and final positions, (b) the direction of her net displacement D as an angle θD south of west.

Background

Topic: Vector Addition and Components in 2-D

This problem tests your ability to break vectors into components, add them algebraically, and find the magnitude and direction of the resultant vector.

Key Terms and Formulas

• Vector components: , (be careful with angles and signs)

• Resultant vector: (add x and y components separately)

• Magnitude:

• Direction: (adjust for quadrant)

Step-by-Step Guidance

1. Break each displacement vector (A, B, C) into x and y components. Assign west and south as negative directions as appropriate.

2. Sum all x-components to get , and all y-components to get .

3. Calculate the magnitude of the net displacement:

4. Find the direction of the net displacement using:

   Be sure to express the angle as south of west, and check the signs/quadrant.

Try solving on your own before revealing the answer!

Q5. The position vector of a particle moving in the x-y plane is given by , where t is in seconds. Calculate (a) the speed of the particle at t = 2.00 s, (b) the direction of its motion at t = 2.00 s, as an angle counterclockwise from the positive x-direction, (c) the direction of its acceleration at t = 2.00 s, as an angle counterclockwise from the positive x-direction.

Background

Topic: 2-D Kinematics, Velocity and Acceleration Vectors

This problem tests your ability to find velocity and acceleration vectors by differentiating the position vector, and then to find their magnitudes and directions at a specific time.

Key Terms and Formulas

• Velocity:

• Acceleration:

• Speed:

• Direction:

Step-by-Step Guidance

1. Differentiate the position vector to find the velocity vector components:

   ,

2. Evaluate and at s.

3. Calculate the speed:

4. Find the direction of the velocity vector at s:

5. Differentiate the velocity vector to find the acceleration vector components, then find its direction at s.

Try solving on your own before revealing the answer!

Q6. A mischievous physics student launches a spit wad vertically downward out of a straw, toward his instructor 3.00 m below. The spit wad splats down after 0.500 s in flight. What is the impact speed?

Background

Topic: 1-D Kinematics with Initial Velocity (Downward Launch)

This problem tests your ability to use kinematic equations to solve for initial velocity and then find the final velocity after a given displacement and time.

Key Terms and Formulas

• Displacement:

• Final velocity:

• Acceleration: m/s2 (downward)

Step-by-Step Guidance

1. Set up the displacement equation for the spit wad's motion:

   Here, m (downward), s.

2. Rearrange to solve for the initial velocity .

3. Once is found, use the velocity equation to find the impact speed:

Try solving on your own before revealing the answer!

Q7. A child drops a toy soldier with a parachute out a window. The soldier descends at a constant speed of 1.50 m/s for 2.00 s, then detaches from the parachute and falls for another 0.900 s in free-fall, hitting the ground exactly 2.90 s after release. What was the height of the window above the ground?

Background

Topic: Piecewise 1-D Kinematics (Uniform Motion and Free-Fall)

This problem involves two phases: (1) constant speed descent, and (2) free-fall. You must calculate the distance for each phase and sum them for the total height.

Key Terms and Formulas

• Constant speed:

• Free-fall: (with as the speed at the moment of detachment)

• Total height:

Step-by-Step Guidance

1. Calculate the distance descended at constant speed:

2. At the moment of detachment, the soldier's speed is 1.50 m/s downward (this is for the free-fall phase).

3. Calculate the distance fallen during free-fall:

   where s.

4. Add and to get the total height.

Try solving on your own before revealing the answer!

Q8. A giant bug starts at t = 0 s with an initial speed of 3.00 m/s from a distance of 4.00 m away from a window, flying directly towards it and speeding up at a constant rate, splattering on the window at t = 0.500 s. What was the magnitude of its acceleration?

Background

Topic: 1-D Kinematics with Constant Acceleration (Solving for Acceleration)

This problem tests your ability to use the kinematic equation for displacement to solve for acceleration, given initial speed, displacement, and time.

Key Terms and Formulas

• Displacement:

• Given: , m, m/s, s

• Unknown:

Step-by-Step Guidance

1. Write the kinematic equation for displacement:

2. Plug in the known values for , , and .

3. Rearrange the equation to solve for .

4. Substitute the numbers and solve for (but do not compute the final value yet).

Try solving on your own before revealing the answer!

Q9. A pellet gun is fired vertically upward at t = 0 s. The pellet hits the ceiling, which is exactly 5.00 m above the muzzle, at t = 0.400 s. With what impact speed does the pellet strike the ceiling?

Background

Topic: 1-D Kinematics with Constant Acceleration (Vertical Motion)

This problem tests your ability to use kinematic equations to solve for initial velocity and then find the final velocity after a given displacement and time, with upward motion against gravity.

Key Terms and Formulas

• Displacement:

• Final velocity:

• Acceleration: m/s2 (upward motion)

Step-by-Step Guidance

1. Set up the displacement equation for the pellet's motion:

   Here, m, s, m/s2.

2. Rearrange to solve for the initial velocity .

3. Once is found, use the velocity equation to find the impact speed:

Try solving on your own before revealing the answer!