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If the velocity of the particle is given by `v=sqrt(x)` and initially particel was at `x=4m` then which of the following are correct.A. at `t=2` sec, the position of the particle is `x=9 m`B. particle acceleration at `t=2` sec is `1 m//s^(2)`C. particle acceleration is `½ ms^(2)` through out the motionD. particle will never go in negative direction from its starting position. |
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Answer» Correct Answer - A::C::D `v=sqrt(x)` `(dx)/(dt)=sqrt(x)impliesint(dx)/x^(1//2)=int dt implies 2sqrt(x)=t+c` But at `t=0,x=4,impliesc=4` `impliesx=((t+4)^(2))/(4)impliesx=((6)^(2))/(4)=(36)/(4)=9` `a=v(dv)/(dx)=sqrt(x)xx(1)/(2sqrt(x))= ½ m//s^(2)`. |
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