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Standard XII

Physics

a-t Graph

A particle is...

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

\frac{1}{2} f_{0} T

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f_{0} T

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\frac{1}{2} f_{0} T^{2}

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f_{0} T^{2}

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Solution

The correct option is A $\frac{1}{2} f_{0} T$
$G i v e n f = f_{0} (1 - \frac{t}{T}) w h e r e f_{0} a n d T = c o n s t a n t A t t = 0 v = 0 W h e n f = 0, t h a n 0 = f_{0} (1 - \frac{t}{T}) T = t - - - - - - (1) w e a c c e l e r a t i o n f = \frac{f i n a l - i n i t i a l v e l o c i t y}{T i m e} \frac{d v}{d t} = f_{0} (1 - \frac{t}{T}) \int d v = \int f_{0} (1 - \frac{t}{T}) d t v = f_{0} (t - \frac{t^{2}}{2 T}) F r o m e q u t a i o n 1 w e g e t t = T v = \frac{1}{2} f_{0} T$

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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 then

The correct option is A 12f0TGivenf=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