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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.
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.
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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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