Physics with Calculus: Revision Worksheet Step-by-Step Guidance

Q1. A car accelerates from rest at 2 m/s². How far does it travel in 5 seconds?

Background

Topic: Kinematics (Motion with Constant Acceleration)

This question tests your understanding of the equations of motion for an object starting from rest and accelerating at a constant rate.

Key formula:

$s = ut + \frac{1}{2} a t^2$

Where:

• $s$ = displacement (distance traveled)

• $u$ = initial velocity (0 m/s, since the car starts from rest)

• $a$ = acceleration (2 m/s²)

• $t$ = time (5 s)

Step-by-Step Guidance

1. Identify the known values: $u = 0$ m/s, $a = 2$ m/s², $t = 5$ s.

2. Write the kinematic equation for displacement: $s = ut + \frac{1}{2} a t^2$.

3. Since $u = 0$, the equation simplifies to $s = \frac{1}{2} a t^2$.

4. Set up the calculation: $s = \frac{1}{2} \times 2 \times (5)^2$.

Try solving on your own before revealing the answer!

Final Answer: 25 meters

$s = \frac{1}{2} \times 2 \times (5)^2 = 1 \times 25 = 25$ meters

The car travels 25 meters in 5 seconds under constant acceleration from rest.

Q2. What is the velocity of the car at the end of 5 seconds?

Background

Topic: Kinematics (Final Velocity with Constant Acceleration)

This question tests your ability to use the kinematic equation for velocity after a period of constant acceleration.

Key formula:

$v = u + a t$

Where:

• $v$ = final velocity

• $u$ = initial velocity (0 m/s)

• $a$ = acceleration (2 m/s²)

• $t$ = time (5 s)

Step-by-Step Guidance

1. Identify the known values: $u = 0$ m/s, $a = 2$ m/s², $t = 5$ s.

2. Write the equation for final velocity: $v = u + a t$.

3. Since $u = 0$, the equation simplifies to $v = a t$.

4. Set up the calculation: $v = 2 \times 5$.

Try solving on your own before revealing the answer!

Final Answer: 10 m/s

$v = 2 \times 5 = 10$ m/s

The car's velocity at the end of 5 seconds is 10 m/s.

Q3. A ball is thrown vertically upwards with a velocity of 15 m/s. How high does it go?

Background

Topic: Kinematics (Projectile Motion - Maximum Height)

This question tests your understanding of how to calculate the maximum height reached by a projectile thrown upwards.

Key formula:

$v^2 = u^2 + 2 a s$

Where:

• $v$ = final velocity (0 m/s at the top)

• $u$ = initial velocity (15 m/s)

• $a$ = acceleration (gravity, $-9.8$ m/s²)

• $s$ = maximum height

Step-by-Step Guidance

1. Identify the known values: $u = 15$ m/s, $v = 0$ m/s, $a = -9.8$ m/s².

2. Write the equation: $v^2 = u^2 + 2 a s$.

3. Set $v = 0$ and solve for $s$: $0 = (15)^2 + 2 \times (-9.8) \times s$.

4. Rearrange to solve for $s$: $s = \frac{-(15)^2}{2 \times (-9.8)}$.

Try solving on your own before revealing the answer!

Final Answer: 11.5 meters

$s = \frac{-(15)^2}{2 \times (-9.8)} = \frac{-225}{-19.6} \approx 11.5$ meters

The ball reaches a maximum height of approximately 11.5 meters.

Q4. What is the time taken for the ball to reach its maximum height?

Background

Topic: Kinematics (Projectile Motion - Time to Maximum Height)

This question tests your ability to calculate the time it takes for a projectile to reach its peak.

Key formula:

$v = u + a t$

Where:

• $v$ = final velocity (0 m/s at the top)

• $u$ = initial velocity (15 m/s)

• $a$ = acceleration (gravity, $-9.8$ m/s²)

• $t$ = time to reach maximum height

Step-by-Step Guidance

1. Identify the known values: $u = 15$ m/s, $v = 0$ m/s, $a = -9.8$ m/s².

2. Write the equation: $v = u + a t$.

3. Set $v = 0$ and solve for $t$: $0 = 15 + (-9.8) t$.

4. Rearrange to solve for $t$: $t = \frac{-15}{-9.8}$.

Try solving on your own before revealing the answer!

Final Answer: 1.53 seconds

$t = \frac{-15}{-9.8} \approx 1.53$ seconds

The ball takes about 1.53 seconds to reach its maximum height.

Q5. Explain the difference between speed and velocity.

Background

Topic: Kinematics (Speed vs. Velocity)

This question tests your understanding of the distinction between two fundamental concepts in motion.

Key Terms:

• Speed: The rate at which an object covers distance; scalar quantity (no direction).

• Velocity: The rate at which an object changes its position; vector quantity (has direction).

Step-by-Step Guidance

1. Define speed as a scalar quantity: it only measures how fast something moves, regardless of direction.

2. Define velocity as a vector quantity: it measures how fast and in what direction something moves.

3. Consider an example: If a car moves 10 meters east in 2 seconds, its speed is $\frac{10}{2} = 5$ m/s, and its velocity is 5 m/s east.

4. Think about cases where direction changes: If the car returns to its starting point, its speed is still positive, but its velocity could be zero.

Try explaining on your own before revealing the answer!

Final Answer:

Speed is a scalar quantity that measures how fast an object moves, while velocity is a vector quantity that measures how fast and in what direction an object moves. Velocity can be zero if the object returns to its starting point, even if speed is positive.

Q6. A train travels 100 km in 2 hours. What is its average speed?

Background

Topic: Kinematics (Average Speed Calculation)

This question tests your ability to calculate average speed using total distance and total time.

Key formula:

$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$

Where:

• Total distance = 100 km

• Total time = 2 hours

Step-by-Step Guidance

1. Identify the known values: Total distance = 100 km, Total time = 2 hours.

2. Write the formula for average speed: $\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$.

3. Set up the calculation: $\text{Average speed} = \frac{100}{2}$.

Try solving on your own before revealing the answer!

Final Answer: 50 km/h

$\text{Average speed} = \frac{100}{2} = 50$ km/h

The train's average speed is 50 km/h.

Q7. If a car moves in a circle at constant speed, is its velocity constant?

Background

Topic: Circular Motion (Velocity vs. Speed)

This question tests your understanding of how velocity changes in circular motion, even if speed is constant.

Key Terms:

• Speed: Magnitude of motion (constant in this case).

• Velocity: Direction and magnitude; direction changes continuously in circular motion.

Step-by-Step Guidance

1. Recall that velocity is a vector, so it depends on both speed and direction.

2. In circular motion, the direction of motion changes at every instant, even if the speed remains constant.

3. Therefore, the velocity is not constant because its direction is always changing.

Try explaining on your own before revealing the answer!

Final Answer:

No, the velocity is not constant because the direction of motion changes continuously in circular motion, even though the speed is constant.

Q8. What is the formula for average velocity?

Background

Topic: Kinematics (Average Velocity)

This question tests your ability to recall and use the formula for average velocity.

Key formula:

$\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}$

Where:

• Total displacement = straight-line distance from start to finish

• Total time = time taken for the journey

Step-by-Step Guidance

1. Recall that average velocity is based on displacement, not total distance.

2. Write the formula: $\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}$.

3. Think about how this differs from average speed, which uses total distance.

Try recalling on your own before revealing the answer!

Final Answer:

$\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}$

Average velocity is calculated using displacement, not total distance.

Q9. A car moves 40 km east and then 30 km west in 2 hours. What is its average velocity?

Background

Topic: Kinematics (Average Velocity Calculation)

This question tests your ability to calculate average velocity using displacement and total time.

Key formula:

$\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}$

Where:

• Total displacement = net change in position (east minus west)

• Total time = 2 hours

Step-by-Step Guidance

1. Calculate the net displacement: $40$km east$- 30$km west$= 10$ km east.

2. Write the formula for average velocity: $\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}$.

3. Set up the calculation: $\text{Average velocity} = \frac{10}{2}$.

Try solving on your own before revealing the answer!

Final Answer: 5 km/h east

$\text{Average velocity} = \frac{10}{2} = 5$ km/h east

The car's average velocity is 5 km/h east.

Q10. What is the difference between distance and displacement?

Background

Topic: Kinematics (Distance vs. Displacement)

This question tests your understanding of the difference between two fundamental concepts in motion.

Key Terms:

• Distance: Total length of the path traveled; scalar quantity.

• Displacement: Straight-line distance from start to finish; vector quantity.

Step-by-Step Guidance

1. Define distance as the total path length traveled, regardless of direction.

2. Define displacement as the shortest straight-line distance from the starting point to the ending point, with direction.

3. Consider an example: If you walk 5 meters east and then 5 meters west, your distance is 10 meters, but your displacement is 0.

Try explaining on your own before revealing the answer!

Final Answer:

Distance is the total length of the path traveled, while displacement is the straight-line distance from the starting point to the ending point, including direction.