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Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone?
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Two cars leave with a time gap of 1 minute from the same point. They m...
Understanding the problem:
We have two cars, which leave from the same point with a time gap of 1 minute. The cars have an acceleration of 0.2 m/s^2. We need to find the time it takes for the distance between the two cars to become three times its initial value when the second car just starts.
Solution:
Let's break down the problem into smaller steps to solve it.
Step 1: Initial conditions:
- Let's assume the initial distance between the two cars is 'd' meters.
- The first car starts and covers a distance 'd' in the first minute.
- The second car starts after a time gap of 1 minute and has not covered any distance yet.
Step 2: Calculating the distance covered by the second car:
- The second car starts with an initial velocity of 0 m/s and accelerates at 0.2 m/s^2.
- Using the equation of motion, s = ut + (1/2)at^2, where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time.
- Let's assume the time taken by the second car to cover the distance 'd' is 't' seconds.
- The distance covered by the second car in time 't' can be calculated as s = 0*t + (1/2)*0.2*t^2 = 0.1t^2 meters.
Step 3: Calculating the distance between the two cars:
- After time 't' seconds, the first car would have covered a distance of 'd + d = 2d' meters.
- The second car would have covered a distance of 0.1t^2 meters.
- The distance between the two cars at this point is 2d - 0.1t^2 meters.
Step 4: Finding the time when the distance becomes three times its initial value:
- We need to find the time 'T' when the distance between the two cars becomes three times its initial value, i.e., 3d.
- Setting up an equation, 2d - 0.1t^2 = 3d.
- Rearranging the equation, 0.1t^2 = d.
- Solving for 't', t^2 = 10d.
- Taking the square root, t = sqrt(10d).
Step 5: Finding the time after the departure of the second car:
- The time after the departure of the second car is the sum of the time taken by the first car (1 minute) and the time 't' calculated in Step 4.
- The total time is T = 1 minute + sqrt(10d) seconds.
Final Answer:
The time after the departure of the second car, when the distance between them becomes three times its value, is 1 minute + sqrt(10d) seconds.
We have two cars, which leave from the same point with a time gap of 1 minute. The cars have an acceleration of 0.2 m/s^2. We need to find the time it takes for the distance between the two cars to become three times its initial value when the second car just starts.
Solution:
Let's break down the problem into smaller steps to solve it.
Step 1: Initial conditions:
- Let's assume the initial distance between the two cars is 'd' meters.
- The first car starts and covers a distance 'd' in the first minute.
- The second car starts after a time gap of 1 minute and has not covered any distance yet.
Step 2: Calculating the distance covered by the second car:
- The second car starts with an initial velocity of 0 m/s and accelerates at 0.2 m/s^2.
- Using the equation of motion, s = ut + (1/2)at^2, where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time.
- Let's assume the time taken by the second car to cover the distance 'd' is 't' seconds.
- The distance covered by the second car in time 't' can be calculated as s = 0*t + (1/2)*0.2*t^2 = 0.1t^2 meters.
Step 3: Calculating the distance between the two cars:
- After time 't' seconds, the first car would have covered a distance of 'd + d = 2d' meters.
- The second car would have covered a distance of 0.1t^2 meters.
- The distance between the two cars at this point is 2d - 0.1t^2 meters.
Step 4: Finding the time when the distance becomes three times its initial value:
- We need to find the time 'T' when the distance between the two cars becomes three times its initial value, i.e., 3d.
- Setting up an equation, 2d - 0.1t^2 = 3d.
- Rearranging the equation, 0.1t^2 = d.
- Solving for 't', t^2 = 10d.
- Taking the square root, t = sqrt(10d).
Step 5: Finding the time after the departure of the second car:
- The time after the departure of the second car is the sum of the time taken by the first car (1 minute) and the time 't' calculated in Step 4.
- The total time is T = 1 minute + sqrt(10d) seconds.
Final Answer:
The time after the departure of the second car, when the distance between them becomes three times its value, is 1 minute + sqrt(10d) seconds.
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Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone?
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Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone?.
Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone?.
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Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone?, a detailed solution for Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone? has been provided alongside types of Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration 0.2 m/s2m/s2. How long after the departure of the second car, does the distance between them becomes equal to three times its value, when the second car just starts? Please solve anyone? theory, EduRev gives you an
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