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Asked by lovemaan5500 | 16 Dec, 2019, 09:35: PM
( dv/dt ) = - v

dv/v = - dt

By integrating above eqn, we get,   ln(v) = -t + C  ..................(1)

where C is constant of integration.

At t = 0, we have v = 10, hence C = ln(10)

Hence eqn.(1) becomes,  ln(v/10) = -t   or    v = 10 e-t ...............(2)

velocity after 2 seconds,   v = 10 e-2 m/s

from eqn.(2), we get, v = dS/dt = 10 e-t ...................(3)

By integrating, S = -10 e-t + C  .....................(4)

where C is constant of integration

Let us assume, at t= 0, distance travelled with retrdation is zero

Hence using eqn.(4),  0 = -10 + C  or   C = 10

Hence eqn.(40 becomes,  S(t) = 10( 1 - e-t )

Hence ditance travelled after 2 seconds,  S = 10 ( 1 - e-2 ) m
Answered by Thiyagarajan K | 17 Dec, 2019, 03:01: PM

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