AHS Physics worksheet on displacement and velocity with six word problems.
Displacement and Velocity Worksheet for AHS Physics, featuring six problems on calculating displacement, average velocity, and echo timing, with space for student name, date, and period.
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Step-by-step solution for: Displacement and Velocity Worksheet
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Show Answer Key & Explanations
Step-by-step solution for: Displacement and Velocity Worksheet
Here are the step-by-step solutions for the problems on your worksheet.
Problem: Calculate total displacement starting at $x = -5\text{ cm}$.
- Walks to $x = 12\text{ cm}$
- Displacement of $-8\text{ cm}$
- Walks to location $x = 7\text{ cm}$
Step-by-step Solution:
Displacement is simply the difference between where you ended and where you started. You do not need to add up all the little steps in the middle; you only care about the final position minus the initial position.
* Final Position ($x_f$): $7\text{ cm}$
* Initial Position ($x_i$): $-5\text{ cm}$
$$ \text{Displacement} (\Delta x) = x_f - x_i $$
$$ \Delta x = 7\text{ cm} - (-5\text{ cm}) $$
$$ \Delta x = 7 + 5 $$
$$ \Delta x = 12\text{ cm} $$
***
Problem: Starts at $150\text{ m}$ north, ends at $1200\text{ m}$ north. Time = $30.0\text{ minutes}$. Find average velocity in $\text{m/s}$.
Step-by-step Solution:
First, find the displacement (change in position), then convert the time into seconds, and finally divide displacement by time.
1. Calculate Displacement:
$$ \Delta x = 1200\text{ m} - 150\text{ m} = 1050\text{ m} \text{ (North)} $$
2. Convert Time to Seconds:
There are $60$ seconds in $1$ minute.
$$ t = 30.0\text{ min} \times 60\text{ s/min} = 1800\text{ s} $$
3. Calculate Average Velocity:
$$ v_{avg} = \frac{\Delta x}{t} $$
$$ v_{avg} = \frac{1050\text{ m}}{1800\text{ s}} $$
$$ v_{avg} \approx 0.583\text{ m/s} $$
Since the movement was North, the velocity is directed North.
***
Statement: "A man walked at an average velocity of $5.2\text{ m/s}$."
Explanation:
Velocity is a vector quantity, which means it must have both a magnitude (speed) and a direction. The statement gives the speed ($5.2\text{ m/s}$) but fails to state the direction (e.g., North, East, or towards the store). Without the direction, it is technically average *speed*, not average *velocity*.
***
Problem: Time = $0.53\text{ hours}$. Average speed = $19\text{ km/h}$. Find displacement.
Step-by-step Solution:
Assuming the bus travels in a straight line from the house to the school, we can use the formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
$$ d = v \times t $$
$$ d = 19\text{ km/h} \times 0.53\text{ h} $$
$$ d = 10.07\text{ km} $$
Rounding to two significant figures (since $19$ and $0.53$ have two sig figs):
$$ d \approx 10\text{ km} $$
The displacement is $10\text{ km}$ in the direction of the school.
***
Problem: A girl practices cross-country but has an average velocity of $0.0\text{ m/s}$. How is this possible?
Explanation:
Average velocity depends on displacement (change in position), not total distance run.
$$ v_{avg} = \frac{\text{Displacement}}{\text{Time}} $$
If her average velocity is zero, her displacement must be zero. This happens if she starts and finishes her practice at the exact same location. For example, if she runs a loop around the track or runs out into the woods and runs back to the start, her net change in position is zero, making her average velocity zero, even though she ran a long distance.
***
Problem: Hiker is $280.5\text{ meters}$ from the wall. Sound speed is $340\text{ m/s}$. How long until she hears the echo?
Step-by-step Solution:
An echo requires the sound to travel from the hiker to the wall AND back to the hiker. Therefore, the total distance the sound travels is double the distance to the wall.
1. Calculate Total Distance:
$$ d_{total} = 2 \times 280.5\text{ m} = 561.0\text{ m} $$
2. Calculate Time:
$$ t = \frac{d}{v} $$
$$ t = \frac{561.0\text{ m}}{340\text{ m/s}} $$
$$ t = 1.65\text{ s} $$
Final Answer:
1. 12 cm
2. 0.583 m/s North
3. The statement is missing the direction. Velocity is a vector and requires a direction (e.g., "5.2 m/s North").
4. 10 km (towards the school)
5. She started and ended at the same location, resulting in zero displacement.
6. 1.65 seconds
1. Total Displacement of the Mouse
Problem: Calculate total displacement starting at $x = -5\text{ cm}$.
- Walks to $x = 12\text{ cm}$
- Displacement of $-8\text{ cm}$
- Walks to location $x = 7\text{ cm}$
Step-by-step Solution:
Displacement is simply the difference between where you ended and where you started. You do not need to add up all the little steps in the middle; you only care about the final position minus the initial position.
* Final Position ($x_f$): $7\text{ cm}$
* Initial Position ($x_i$): $-5\text{ cm}$
$$ \text{Displacement} (\Delta x) = x_f - x_i $$
$$ \Delta x = 7\text{ cm} - (-5\text{ cm}) $$
$$ \Delta x = 7 + 5 $$
$$ \Delta x = 12\text{ cm} $$
***
2. Average Velocity of the Bicyclist
Problem: Starts at $150\text{ m}$ north, ends at $1200\text{ m}$ north. Time = $30.0\text{ minutes}$. Find average velocity in $\text{m/s}$.
Step-by-step Solution:
First, find the displacement (change in position), then convert the time into seconds, and finally divide displacement by time.
1. Calculate Displacement:
$$ \Delta x = 1200\text{ m} - 150\text{ m} = 1050\text{ m} \text{ (North)} $$
2. Convert Time to Seconds:
There are $60$ seconds in $1$ minute.
$$ t = 30.0\text{ min} \times 60\text{ s/min} = 1800\text{ s} $$
3. Calculate Average Velocity:
$$ v_{avg} = \frac{\Delta x}{t} $$
$$ v_{avg} = \frac{1050\text{ m}}{1800\text{ s}} $$
$$ v_{avg} \approx 0.583\text{ m/s} $$
Since the movement was North, the velocity is directed North.
***
3. What is wrong with the statement?
Statement: "A man walked at an average velocity of $5.2\text{ m/s}$."
Explanation:
Velocity is a vector quantity, which means it must have both a magnitude (speed) and a direction. The statement gives the speed ($5.2\text{ m/s}$) but fails to state the direction (e.g., North, East, or towards the store). Without the direction, it is technically average *speed*, not average *velocity*.
***
4. Displacement of the School Bus
Problem: Time = $0.53\text{ hours}$. Average speed = $19\text{ km/h}$. Find displacement.
Step-by-step Solution:
Assuming the bus travels in a straight line from the house to the school, we can use the formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
$$ d = v \times t $$
$$ d = 19\text{ km/h} \times 0.53\text{ h} $$
$$ d = 10.07\text{ km} $$
Rounding to two significant figures (since $19$ and $0.53$ have two sig figs):
$$ d \approx 10\text{ km} $$
The displacement is $10\text{ km}$ in the direction of the school.
***
5. Zero Average Velocity in Cross-Country
Problem: A girl practices cross-country but has an average velocity of $0.0\text{ m/s}$. How is this possible?
Explanation:
Average velocity depends on displacement (change in position), not total distance run.
$$ v_{avg} = \frac{\text{Displacement}}{\text{Time}} $$
If her average velocity is zero, her displacement must be zero. This happens if she starts and finishes her practice at the exact same location. For example, if she runs a loop around the track or runs out into the woods and runs back to the start, her net change in position is zero, making her average velocity zero, even though she ran a long distance.
***
6. Echo in the Canyon
Problem: Hiker is $280.5\text{ meters}$ from the wall. Sound speed is $340\text{ m/s}$. How long until she hears the echo?
Step-by-step Solution:
An echo requires the sound to travel from the hiker to the wall AND back to the hiker. Therefore, the total distance the sound travels is double the distance to the wall.
1. Calculate Total Distance:
$$ d_{total} = 2 \times 280.5\text{ m} = 561.0\text{ m} $$
2. Calculate Time:
$$ t = \frac{d}{v} $$
$$ t = \frac{561.0\text{ m}}{340\text{ m/s}} $$
$$ t = 1.65\text{ s} $$
Final Answer:
1. 12 cm
2. 0.583 m/s North
3. The statement is missing the direction. Velocity is a vector and requires a direction (e.g., "5.2 m/s North").
4. 10 km (towards the school)
5. She started and ended at the same location, resulting in zero displacement.
6. 1.65 seconds
Parent Tip: Review the logic above to help your child master the concept of velocity problems worksheet.
