17 Straight Line Motion Classwork Answers Overview
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Straight line motion is a fundamental concept in physics, and understanding how objects move in a straight line can help us analyze and predict their behavior. In a recent classwork assignment, students were tasked with solving problems related to straight line motion. Here are the answers to 17 straight line motion classwork problems:
1. A car travels at a constant speed of 60 mph for 2 hours. How far does the car travel?
Answer: The distance traveled by the car can be calculated using the formula distance = speed x time. In this case, distance = 60 mph x 2 hours = 120 miles.
2. An object is dropped from a height of 100 meters. How long does it take for the object to hit the ground?
Answer: The time it takes for an object to fall from a certain height can be calculated using the formula time = sqrt(2h/g), where h is the height and g is the acceleration due to gravity (9.8 m/s^2). In this case, time = sqrt(2 x 100 m / 9.8 m/s^2) ≈ 4.52 seconds.
3. A train accelerates from rest at a rate of 2 m/s^2 for 10 seconds. What is the final velocity of the train?
Answer: The final velocity of the train can be calculated using the formula final velocity = initial velocity + acceleration x time. Since the train starts from rest, the initial velocity is 0 m/s. Therefore, final velocity = 0 m/s + 2 m/s^2 x 10 seconds = 20 m/s.
4. A ball is thrown vertically upwards with an initial velocity of 20 m/s. How high does the ball go before falling back down?
Answer: The height reached by the ball can be calculated using the formula height = (initial velocity^2) / (2 x gravity). In this case, height = (20 m/s)^2 / (2 x 9.8 m/s^2) = 20.41 meters.
5. A car decelerates from 30 m/s to 10 m/s in 5 seconds. What is the car’s acceleration?
Answer: The acceleration of the car can be calculated using the formula acceleration = (final velocity – initial velocity) / time. In this case, acceleration = (10 m/s – 30 m/s) / 5 seconds = -4 m/s^2 (negative sign indicates deceleration).
6. A rocket is launched with an initial velocity of 100 m/s. If the rocket accelerates at a rate of 5 m/s^2, how long does it take for the rocket to reach a velocity of 200 m/s?
Answer: The time it takes for the rocket to reach a certain velocity can be calculated using the formula time = (final velocity – initial velocity) / acceleration. In this case, time = (200 m/s – 100 m/s) / 5 m/s^2 = 20 seconds.
7. A stone is thrown horizontally from a cliff with an initial velocity of 15 m/s. How far does the stone travel horizontally before hitting the ground?
Answer: The horizontal distance traveled by the stone can be calculated using the formula distance = velocity x time. Since the stone is thrown horizontally, the initial vertical velocity does not affect the horizontal distance. Therefore, distance = 15 m/s x time. Additional information is needed to calculate the total distance traveled.
8. A particle moves along a straight line with a velocity of 5 m/s. If the acceleration of the particle is 2 m/s^2, what is the particle’s displacement after 3 seconds?
Answer: The displacement of the particle can be calculated using the formula displacement = initial velocity x time + (1/2) x acceleration x (time)^2. In this case, displacement = 5 m/s x 3 seconds + (1/2) x 2 m/s^2 x (3 seconds)^2 = 25.5 meters.
9. A car accelerates uniformly from rest to a velocity of 20 m/s in 4 seconds. What is the acceleration of the car?
Answer: The acceleration of the car can be calculated using the formula acceleration = (final velocity – initial velocity) / time. In this case, acceleration = (20 m/s – 0 m/s) / 4 seconds = 5 m/s^2.
10. A stone is dropped from a height of 50 meters. How long does it take for the stone to hit the ground?
Answer: The time it takes for the stone to fall from a certain height can be calculated using the formula time = sqrt(2h/g), where h is the height and g is the acceleration due to gravity (9.8 m/s^2). In this case, time = sqrt(2 x 50 m / 9.8 m/s^2) ≈ 3.19 seconds.
11. A car traveling at 60 mph stops in a distance of 50 meters. What is the car’s deceleration?
Answer: The deceleration of the car can be calculated using the formula deceleration = (final velocity^2 – initial velocity^2) / (2 x distance). In this case, final velocity = 0 m/s (car stops), so deceleration = (0 m/s^2 – (60 mph)^2) / (2 x 50 meters) ≈ -8.46 m/s^2.
12. A truck accelerates from rest at a rate of 3 m/s^2 for 8 seconds. How far does the truck travel during this time?
Answer: The distance traveled by the truck can be calculated using the formula distance = (initial velocity x time) + (1/2) x acceleration x (time)^2. Since the truck starts from rest, the initial velocity is 0 m/s. Therefore, distance = (0 m/s x 8 seconds) + (1/2) x 3 m/s^2 x (8 seconds)^2 = 96 meters.
13. A ball is thrown vertically upwards with an initial velocity of 30 m/s. How high does the ball go before falling back down?
Answer: The height reached by the ball can be calculated using the formula height = (initial velocity^2) / (2 x gravity). In this case, height = (30 m/s)^2 / (2 x 9.8 m/s^2) ≈ 46.94 meters.
14. A car decelerates from 40 m/s to 20 m/s in 6 seconds. What is the car’s acceleration?
Answer: The acceleration of the car can be calculated using the formula acceleration = (final velocity – initial velocity) / time. In this case, acceleration = (20 m/s – 40 m/s) / 6 seconds = -3.33 m/s^2 (negative sign indicates deceleration).
15. A rocket is launched with an initial velocity of 200 m/s. If the rocket accelerates at a rate of 10 m/s^2, how long does it take for the rocket to reach a velocity of 500 m/s?
Answer: The time it takes for the rocket to reach a certain velocity can be calculated using the formula time = (final velocity – initial velocity) / acceleration. In this case, time = (500 m/s – 200 m/s) / 10 m/s^2 = 30 seconds.
16. A stone is thrown horizontally from a cliff with an initial velocity of 25 m/s. How far does the stone travel horizontally before hitting the ground?
Answer: The horizontal distance traveled by the stone can be calculated using the formula distance = velocity x time. Since the stone is thrown horizontally, the initial vertical velocity does not affect the horizontal distance. Therefore, distance = 25 m/s x time. Additional information is needed to calculate the total distance traveled.
17. A particle moves along a straight line with a velocity of 10 m/s. If the acceleration of the particle is -3 m/s^2, what is the particle’s displacement after 2 seconds?
Answer: The displacement of the particle can be calculated using the formula displacement = initial velocity x time + (1/2) x acceleration x (time)^2. In this case, displacement = 10 m/s x 2 seconds + (1/2) x -3 m/s^2 x (2 seconds)^2 = 16 meters.
In conclusion, these 17 straight line motion classwork problems cover a range of scenarios involving objects moving in a straight line. By understanding and applying the basic concepts of velocity, acceleration, time, and distance, students can effectively analyze and solve problems related to straight line motion. Practicing these types of problems can help reinforce key physics principles and improve problem-solving skills in this area of study.
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