뒤로Physics with Calculus: Step-by-Step Guidance for Kinematics and Vector Problems
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Q1. If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattentive period?
Background
Topic: Kinematics – Constant Velocity
This question tests your understanding of the relationship between speed, time, and distance when moving at a constant velocity.
Key Terms and Formulas
Speed (): The rate at which an object covers distance.
Distance (): The total length of the path traveled.
Time (): The duration of travel.
Key formula:
Step-by-Step Guidance
First, convert the speed from km/h to m/s to match the SI units for distance and time.
Recall that , so multiply the speed by this conversion factor.
Once you have the speed in m/s, use the formula to set up the calculation for distance.
Plug in the converted speed and the given time (4.0 s) into the formula, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 80 m
First, convert 72 km/h to m/s: .
Then, .
So, you travel 80 meters during the 4 seconds of inattention.
Q2. An airplane increases its speed at the average rate of 15 m/s2. How much time does it take to increase its speed from 100 m/s to 160 m/s?
Background
Topic: Kinematics – Constant Acceleration
This question tests your ability to use the equations of motion for constant acceleration to relate initial and final velocities, acceleration, and time.
Key Terms and Formulas
Initial velocity (): The starting speed of the airplane.
Final velocity (): The speed after accelerating.
Acceleration (): The rate of change of velocity.
Time (): The duration of acceleration.
Key formula:
Step-by-Step Guidance
Identify the known values: , , .
Write the equation for final velocity: .
Rearrange the equation to solve for time: .
Substitute the known values into the equation, but do not calculate the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 4.0 s
The airplane takes 4.0 seconds to increase its speed from 100 m/s to 160 m/s at the given acceleration.
Q3. (Calculus Based) A particle's position is given by m. What is the velocity and acceleration of the particle when s?
Background
Topic: Calculus in Kinematics – Position, Velocity, and Acceleration
This question tests your ability to use derivatives to find velocity and acceleration from a position function.
Key Terms and Formulas
Position (): The location of the particle as a function of time.
Velocity (): The first derivative of position with respect to time.
Acceleration (): The second derivative of position with respect to time, or the first derivative of velocity.
Key formulas:
Step-by-Step Guidance
Write the position function: .
Find the velocity function by differentiating with respect to .
Find the acceleration function by differentiating with respect to .
Substitute s into the expressions for and , but do not compute the final values yet.
Try solving on your own before revealing the answer!
Final Answer: ,
Velocity:
At s:
Acceleration:
At s:
Q4. (Calculus Based) A car starting from rest accelerates at a constant 2.0 m/s2 for 10 s. It then travels with the constant speed it has achieved for another 10 s. Then it finally slows to a stop with constant acceleration of magnitude 2.0 m/s2. How far does it travel after starting?
Background
Topic: Kinematics – Piecewise Motion (Acceleration, Constant Speed, Deceleration)
This question tests your ability to analyze motion in stages, using kinematic equations for each segment and summing the total distance.
Key Terms and Formulas
Acceleration (): The rate of change of velocity.
Constant velocity: Motion with zero acceleration.
Kinematic equations for constant acceleration:
Step-by-Step Guidance
For the first segment, use to find the distance covered during acceleration.
Find the velocity at the end of the first segment using ; this will be the constant speed for the next segment.
For the second segment, use to find the distance covered at constant speed.
For the third segment (deceleration), use the kinematic equation (with negative) to find the distance required to come to rest.
Add the distances from all three segments to set up the total distance traveled, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 400 m
First segment:
Velocity after first segment:
Second segment:
Third segment:
Total distance:
Q5. A car increases its forward velocity uniformly from 40 m/s to 80 m/s while traveling a distance of 200 m. What is its acceleration during this time?
Background
Topic: Kinematics – Constant Acceleration (No Time Given)
This question tests your ability to use kinematic equations to solve for acceleration when initial and final velocities and displacement are known, but time is not given.
Key Terms and Formulas
Initial velocity (), final velocity (), displacement (), acceleration ().
Kinematic equation (no time):
Step-by-Step Guidance
Identify the known values: , , .
Write the kinematic equation: .
Rearrange the equation to solve for acceleration: .
Substitute the known values into the equation, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 12 m/s2
The car's acceleration is 12 m/s2.
Q6. A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s2. How far does the car travel after the driver put on the brakes until it comes to a stop?
Background
Topic: Kinematics – Deceleration to Rest
This question tests your ability to use kinematic equations to find stopping distance when initial speed and acceleration are known.
Key Terms and Formulas
Initial velocity (), final velocity (), acceleration (), displacement ().
Kinematic equation:
Step-by-Step Guidance
Identify the known values: , , (negative because it's slowing down).
Write the kinematic equation: .
Rearrange to solve for : .
Substitute the known values into the equation, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 146 m
The car travels 146 meters before coming to a stop.
Q7. A bullet shot straight up returns to its starting point in 10 s. What is the initial speed of the bullet, assuming negligible air resistance?
Background
Topic: Kinematics – Vertical Projectile Motion
This question tests your understanding of the symmetry of projectile motion and the use of kinematic equations for vertical motion under gravity.
Key Terms and Formulas
Initial velocity (), time of flight (), acceleration due to gravity ().
For an object thrown vertically upward and returning to the same point:
Step-by-Step Guidance
Recognize that the total time up and down is s.
Use the formula to relate initial speed and total time.
Rearrange to solve for : .
Substitute and s, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 49 m/s
The initial speed of the bullet is 49 m/s.
Q8. A ball is thrown downward from the top of a building with an initial speed of 25 m/s. It strikes the ground after 2.0 s. How high is the building, assuming negligible air resistance?
Background
Topic: Kinematics – Vertical Motion with Initial Downward Velocity
This question tests your ability to use kinematic equations for motion under gravity with an initial velocity directed downward.
Key Terms and Formulas
Initial velocity (), time (), acceleration due to gravity (), displacement ().
Kinematic equation:
Step-by-Step Guidance
Identify the known values: (downward), s, .
Write the equation for displacement: .
Substitute the known values into the equation, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 70 m
The building is approximately 70 meters high.
Q9. Displacement vector A is 75 cm long and points at 30° above the +x-axis. Displacement vector B is 25 cm long and points along the -x-axis. Displacement vector C is 40 cm long and points at 45° below the -x-axis. Determine the magnitude and direction of the resultant of these three vectors.
Background
Topic: Vectors – Addition and Components
This question tests your ability to break vectors into components, sum them, and find the magnitude and direction of the resultant vector.
Key Terms and Formulas
Vector components: ,
Resultant vector: ,
Magnitude:
Direction:
Step-by-Step Guidance
Break each vector into its x and y components using trigonometry.
Sum all x-components to get and all y-components to get .
Set up the formula for the magnitude of the resultant: .
Set up the formula for the direction: , but do not compute the final values yet.
Try solving on your own before revealing the answer!
Final Answer: 15 cm at 38° above the +x-axis
After breaking all vectors into components and summing, the resultant has a magnitude of 15 cm and points 38° above the +x-axis.
Q10. (Calculus Based) The velocity of a particle is given by m/s. Find the magnitude and direction of the acceleration when s.
Background
Topic: Calculus in Kinematics – Velocity and Acceleration Vectors
This question tests your ability to find acceleration as the derivative of velocity and to compute the magnitude and direction of a vector.
Key Terms and Formulas
Velocity vector , acceleration vector
Magnitude:
Direction:
Step-by-Step Guidance
Differentiate each component of the velocity vector with respect to to find and .
Evaluate and at s.
Set up the formula for the magnitude: .
Set up the formula for the direction: , but do not compute the final values yet.
Try solving on your own before revealing the answer!
Final Answer: 11.3 m/s2, -23.4°
; at ,
Magnitude:
Direction:
Q11. A girl throws a rock horizontally, with a velocity of 10 m/s, from a bridge. It falls 20 m to the water below. How far does the rock travel horizontally before striking the water, assuming negligible air resistance?
Background
Topic: Projectile Motion – Horizontal Launch
This question tests your ability to analyze projectile motion, separating horizontal and vertical components.
Key Terms and Formulas
Horizontal velocity (), vertical displacement (), time of flight ().
Vertical motion:
Horizontal distance:
Step-by-Step Guidance
Use the vertical motion equation to solve for the time it takes to fall 20 m.
Set up the equation: and solve for .
Once you have , use the horizontal motion equation to set up the calculation for horizontal distance, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 20 m
Time to fall: s
Horizontal distance: m
The rock travels 20 meters horizontally before hitting the water.
Q12. A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and the fence is 2.0 m high. The ball just clears the fence while still traveling upwards and experiences no significant air resistance. How far is the child from the fence?
Background
Topic: Projectile Motion – Two-Dimensional Kinematics
This question tests your ability to analyze projectile motion with both horizontal and vertical components, and to solve for distance when the projectile reaches a certain height.
Key Terms and Formulas
Initial speed (), launch angle (), initial height (), final height (), horizontal distance ().
Vertical motion:
Horizontal motion:
,
Step-by-Step Guidance
Calculate and using the given initial speed and angle.
Set up the vertical motion equation for when the ball is at the height of the fence ( m).
Solve the vertical equation for the time when the ball reaches m (while still rising).
Use this value of in the horizontal motion equation to set up the calculation for , but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 1.6 m
Solving the quadratic for gives s (while rising).
Horizontal distance: m
The child is 1.6 meters from the fence.
Q13. A child throws a ball with an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and experiences no appreciable air resistance as it moves.
(a) How far from where the child is standing does the ball hit the ground?
(b) How long is the ball in flight before it hits the ground?
(c) What is the magnitude of the ball's velocity just before it hits the ground?
Background
Topic: Projectile Motion – Two-Dimensional Kinematics with Initial Height
This question tests your ability to analyze projectile motion, including finding range, time of flight, and final velocity when the projectile starts above ground level.
Key Terms and Formulas
Initial speed (), launch angle (), initial height (), horizontal distance (), time of flight (), final velocity ().
Vertical motion:
Horizontal motion:
Final velocity components: ,
Magnitude:
Step-by-Step Guidance
Calculate and using the initial speed and angle.
Set up the vertical motion equation for when the ball hits the ground ().
Solve the quadratic equation for (time of flight).
Use to set up the calculation for horizontal distance and for the final velocity components, but do not compute the final values yet.
Try solving on your own before revealing the answer!
Final Answers:
(a) 7.5 m
(b) 1.2 s
(c) 9.1 m/s
Solving the quadratic for gives s. The horizontal distance is m. The final velocity magnitude is m/s.