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A bus starts from rest with an acceleration of 1ms-2. A man who is 48m behind the bus starts with a uniform velocity of 10 ms-1. The minimum time after which the man will catch the bus is?
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A bus starts from rest with an acceleration of 1ms-2. A man who is 48m...
Problem: A bus starts from rest with an acceleration of 1ms-2. A man who is 48m behind the bus starts with a uniform velocity of 10 ms-1. The minimum time after which the man will catch the bus is?
Solution:
To find the minimum time after which the man will catch the bus, we need to use the equations of motion.
Step 1: Write down the known values:
Initial velocity of the man, u = 10 ms-1
Acceleration of the bus, a = 1 ms-2
Initial distance between the bus and the man, s = 48 m
Step 2: Determine the equation to use:
Since the bus is accelerating, the equation to use is:
s = ut + 1/2 at^2
Step 3: Substitute the known values into the equation:
48 = 10t + 1/2(1)t^2
Step 4: Simplify the equation:
1/2t^2 + 10t - 48 = 0
Step 5: Solve for t using the quadratic formula:
t = (-b ± √b^2 - 4ac) / 2a
where a = 1/2, b = 10, and c = -48
Substituting these values, we get:
t = (-10 ± √(10^2 - 4(1/2)(-48))) / 2(1/2)
t = (-10 ± √676) / 1
t = (-10 ± 26) / 1
We can ignore the negative solution, so:
t = (16) / 1
Step 6: Check the answer:
The time taken for the man to catch the bus is 16 seconds. We can verify this by calculating the distance traveled by the bus during this time:
s = ut + 1/2 at^2
s = 0 + 1/2(1)(16)^2
s = 128 meters
Since the distance traveled by the man in 16 seconds is also 160 meters (10 m/s x 16 s), he will catch the bus exactly at this point.
Answer: The minimum time after which the man will catch the bus is 16 seconds.
Solution:
To find the minimum time after which the man will catch the bus, we need to use the equations of motion.
Step 1: Write down the known values:
Initial velocity of the man, u = 10 ms-1
Acceleration of the bus, a = 1 ms-2
Initial distance between the bus and the man, s = 48 m
Step 2: Determine the equation to use:
Since the bus is accelerating, the equation to use is:
s = ut + 1/2 at^2
Step 3: Substitute the known values into the equation:
48 = 10t + 1/2(1)t^2
Step 4: Simplify the equation:
1/2t^2 + 10t - 48 = 0
Step 5: Solve for t using the quadratic formula:
t = (-b ± √b^2 - 4ac) / 2a
where a = 1/2, b = 10, and c = -48
Substituting these values, we get:
t = (-10 ± √(10^2 - 4(1/2)(-48))) / 2(1/2)
t = (-10 ± √676) / 1
t = (-10 ± 26) / 1
We can ignore the negative solution, so:
t = (16) / 1
Step 6: Check the answer:
The time taken for the man to catch the bus is 16 seconds. We can verify this by calculating the distance traveled by the bus during this time:
s = ut + 1/2 at^2
s = 0 + 1/2(1)(16)^2
s = 128 meters
Since the distance traveled by the man in 16 seconds is also 160 meters (10 m/s x 16 s), he will catch the bus exactly at this point.
Answer: The minimum time after which the man will catch the bus is 16 seconds.
Community Answer
A bus starts from rest with an acceleration of 1ms-2. A man who is 48m...
In BUS FRAME-
velocity of man = 10 m/s
s = 48 m
Acceleration of man with respect to bus = Acceleration of man - Acceleration of bus = 0 - (1) = -1 m/s
Applying second equation of motion,
s = ut + 1/2 at2
48 = 10t - 1/2*t2
solving we get, t = 12 sec or t = 8 sec
therefore the minimum time is 8 seconds.
velocity of man = 10 m/s
s = 48 m
Acceleration of man with respect to bus = Acceleration of man - Acceleration of bus = 0 - (1) = -1 m/s
Applying second equation of motion,
s = ut + 1/2 at2
48 = 10t - 1/2*t2
solving we get, t = 12 sec or t = 8 sec
therefore the minimum time is 8 seconds.
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A bus starts from rest with an acceleration of 1ms-2. A man who is 48m behind the bus starts with a uniform velocity of 10 ms-1. The minimum time after which the man will catch the bus is?
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A bus starts from rest with an acceleration of 1ms-2. A man who is 48m behind the bus starts with a uniform velocity of 10 ms-1. The minimum time after which the man will catch the bus is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A bus starts from rest with an acceleration of 1ms-2. A man who is 48m behind the bus starts with a uniform velocity of 10 ms-1. The minimum time after which the man will catch the bus is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bus starts from rest with an acceleration of 1ms-2. A man who is 48m behind the bus starts with a uniform velocity of 10 ms-1. The minimum time after which the man will catch the bus is?.
A bus starts from rest with an acceleration of 1ms-2. A man who is 48m behind the bus starts with a uniform velocity of 10 ms-1. The minimum time after which the man will catch the bus is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A bus starts from rest with an acceleration of 1ms-2. A man who is 48m behind the bus starts with a uniform velocity of 10 ms-1. The minimum time after which the man will catch the bus is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bus starts from rest with an acceleration of 1ms-2. A man who is 48m behind the bus starts with a uniform velocity of 10 ms-1. The minimum time after which the man will catch the bus is?.
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