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.
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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