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A body is moving with uniform velocity of 8m/s.when the body just crossed another body ,the second one starts and moves with uniform acceleration of 4m/s2 the time after which two bodies meet , will be how much?
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A body is moving with uniform velocity of 8m/s.when the body just cros...
Given:
Velocity of the first body (v1) = 8 m/s
Acceleration of the second body (a2) = 4 m/s^2
To Find:
Time taken for the two bodies to meet (t)
Concept:
When two bodies are moving in the same direction, the relative velocity between them is given by the difference in their velocities.
Calculation:
Let's assume that the initial distance between the two bodies is 'd' and the time taken for them to meet is 't'.
The distance covered by the first body in time 't' is given by:
Distance1 = v1 * t
The distance covered by the second body in time 't' is given by:
Distance2 = (1/2) * a2 * t^2 (using the equation of motion for uniformly accelerating body)
Since the second body starts moving when the first body has already covered the distance 'd', the equation becomes:
Distance2 = d + v1 * t + (1/2) * a2 * t^2
Since the two bodies meet when they cover the same distance, we can equate Distance1 and Distance2:
v1 * t = d + v1 * t + (1/2) * a2 * t^2
Simplifying the equation:
0 = d + (1/2) * a2 * t^2
Rearranging the equation:
(1/2) * a2 * t^2 = -d
Multiplying both sides of the equation by 2 to eliminate the fraction:
a2 * t^2 = -2d
Dividing both sides of the equation by a2 to solve for t^2:
t^2 = (-2d) / a2
Taking the square root of both sides to solve for t:
t = sqrt((-2d) / a2)
Since the distance 'd' is not given in the question, it is not possible to calculate the exact time taken for the two bodies to meet. However, the given equation provides the relationship between the distance and time for the two bodies to meet.
Velocity of the first body (v1) = 8 m/s
Acceleration of the second body (a2) = 4 m/s^2
To Find:
Time taken for the two bodies to meet (t)
Concept:
When two bodies are moving in the same direction, the relative velocity between them is given by the difference in their velocities.
Calculation:
Let's assume that the initial distance between the two bodies is 'd' and the time taken for them to meet is 't'.
The distance covered by the first body in time 't' is given by:
Distance1 = v1 * t
The distance covered by the second body in time 't' is given by:
Distance2 = (1/2) * a2 * t^2 (using the equation of motion for uniformly accelerating body)
Since the second body starts moving when the first body has already covered the distance 'd', the equation becomes:
Distance2 = d + v1 * t + (1/2) * a2 * t^2
Since the two bodies meet when they cover the same distance, we can equate Distance1 and Distance2:
v1 * t = d + v1 * t + (1/2) * a2 * t^2
Simplifying the equation:
0 = d + (1/2) * a2 * t^2
Rearranging the equation:
(1/2) * a2 * t^2 = -d
Multiplying both sides of the equation by 2 to eliminate the fraction:
a2 * t^2 = -2d
Dividing both sides of the equation by a2 to solve for t^2:
t^2 = (-2d) / a2
Taking the square root of both sides to solve for t:
t = sqrt((-2d) / a2)
Since the distance 'd' is not given in the question, it is not possible to calculate the exact time taken for the two bodies to meet. However, the given equation provides the relationship between the distance and time for the two bodies to meet.
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A body is moving with uniform velocity of 8m/s.when the body just crossed another body ,the second one starts and moves with uniform acceleration of 4m/s2 the time after which two bodies meet , will be how much?
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A body is moving with uniform velocity of 8m/s.when the body just crossed another body ,the second one starts and moves with uniform acceleration of 4m/s2 the time after which two bodies meet , will be how much? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A body is moving with uniform velocity of 8m/s.when the body just crossed another body ,the second one starts and moves with uniform acceleration of 4m/s2 the time after which two bodies meet , will be how much? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is moving with uniform velocity of 8m/s.when the body just crossed another body ,the second one starts and moves with uniform acceleration of 4m/s2 the time after which two bodies meet , will be how much?.
A body is moving with uniform velocity of 8m/s.when the body just crossed another body ,the second one starts and moves with uniform acceleration of 4m/s2 the time after which two bodies meet , will be how much? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A body is moving with uniform velocity of 8m/s.when the body just crossed another body ,the second one starts and moves with uniform acceleration of 4m/s2 the time after which two bodies meet , will be how much? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is moving with uniform velocity of 8m/s.when the body just crossed another body ,the second one starts and moves with uniform acceleration of 4m/s2 the time after which two bodies meet , will be how much?.
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