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A bus accelerates at 2m/s2 for first 10 second after the start and retards at 1m/s2 for next 10vsecond.bFind the total distance covered by the bus?
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A bus accelerates at 2m/s2 for first 10 second after the start and ret...
Problem Statement: A bus accelerates at 2m/s2 for first 10 seconds after the start and retards at 1m/s2 for next 10 seconds. Find the total distance covered by the bus?

Solution:
To solve this problem, we need to use the basic formula of kinematics:

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.

Step 1: Calculate the distance covered during acceleration phase:

During the acceleration phase, the bus has an initial velocity of 0 m/s and an acceleration of 2 m/s2. Therefore,

s = 0(10) + 1/2 (2)(10)^2 = 100 m

The bus covers a distance of 100 meters during the acceleration phase.

Step 2: Calculate the distance covered during retardation phase:

During the retardation phase, the bus has an initial velocity of 20 m/s (which is its final velocity at the end of acceleration phase) and a retardation of 1 m/s2. Therefore,

s = 20(10) + 1/2 (-1)(10)^2 = 200 m

The bus covers a distance of 200 meters during the retardation phase.

Step 3: Calculate the total distance covered:

The total distance covered by the bus is the sum of the distance covered during acceleration and retardation phases.

Therefore, Total distance covered = 100 + 200 = 300 m

Hence, the total distance covered by the bus is 300 meters.
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A bus accelerates at 2m/s2 for first 10 second after the start and ret...
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