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A point moves in a straight …

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A point moves in a straight line under the retardation av^2, where a is positive constant and v is speed. If initial velocity is u, distance covered in t second is
  • 1 answers

Gaurav Seth 6 years ago

 

acceleration = -av^2. (We are assuming that a is a constant and v represents velocity here)

Now, acceleration = dv/dt

Hence, dv/dt = -av^2

dv/v^2 = -adt

Integrating from t = 0, v = u to t = t and v =v 

 

[1/u-1/v] = -at

1/v = 1/u+ at

1/v = (1+uat)/u 

v = u/(1+uat)

Now v = dx/dt

dx/dt = u/(1+uat)

dx = u/(1+uat) dt

Integrating from x = 0, t = 0 to x = x, t = t

x = u/ua * ln|1+uat| 

x = ln|1+uat|/a 

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