A particle is moving along x-axis has acceleration $f$ at time $t$ , given by $f = f_{0} (1 - \frac{t}{T})$ , where $f_{0}$ and $T$ are constants. The particle at $t = 0$ has zero velocity. In the time interval between $t = 0$ and the instant when $f = 0$ , the velocity( $v_{a}$ ) of the particle is then

Question

A particle is moving along x-axis has acceleration

f

at time

t

, given by

f = f_{0} (1 - \frac{t}{T})

, where

f_{0}

and

T

are constants. The particle at

t = 0

has zero velocity. In the time interval between

t = 0

and the instant when

f = 0

, the velocity(

v_{a}

) of the particle is then

Toppr Admin · Accepted Answer

Givenf-f0-1-x2212-tT-wheref0andT-constantAtt-0v-0Whenf-0-than0-f0-1-x2212-tT-T-t-x2212-x2212-x2212-x2212-x2212-x2212-1-weaccelerationf-final-x2212-initialvelocityTimedvdt-f0-1-x2212-tT-x222B-dv-x222B-f0-1-x2212-tT-dtv-f0-t-x2212-t22T-Fromequtaion1wegett-Tv-12f0T

A particle is moving along x-axis has acceleration f at time t, given by f=f0(1−tT), where f0 and T are constants. The particle at t=0 has zero velocity. In the time interval between t=0 and the instant when f=0, the velocity(va) of the particle is then12f0Tf0T12f0T2f0T2

Givenf=f0(1−tT)wheref0andT=constantAtt=0v=0Whenf=0,than0=f0(1−tT)T=t−−−−−−(1)weaccelerationf=final−initialvelocityTimedvdt=f0(1−tT)∫dv=∫f0(1−tT)dtv=f0(t−t22T)Fromequtaion1wegett=Tv=12f0T