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Standard IX

Physics

F=ma

A constant fo...

Question

A constant force $F$ acts on a particle of mass 1 kg moving with a velocity $v$ , for one second. The distance moved in that time is :

0

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\frac{F}{2}

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2 F

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\frac{v}{2}

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v + \frac{F}{2}

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Solution

The correct option is D $v + \frac{F}{2}$
Given, $m = 1 k g$ and $t = 1 s$
We know that, $F = m a$
$a = \frac{F}{m} = \frac{F}{1} = F$
From equation of motion,
$s = u t + \frac{1}{2} a t^{2} = v + \frac{1}{2} F (1)^{2} = v + \frac{F}{2}$

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A constant force F acts on a particle of mass 1 kg moving with a velocity v, for one second. The distance moved in that time is :

The correct option is D v+F2Given, m=1kg and t=1sWe know that, F=maa=Fm=F1=FFrom equation of motion,s=ut+12at2=v+12F(1)2=v+F2

A constant force $F$ acts on a particle of mass 1 kg moving with a velocity $v$ , for one second. The distance moved in that time is :

The correct option is D $v + \frac{F}{2}$
Given, $m = 1 k g$ and $t = 1 s$
We know that, $F = m a$
$a = \frac{F}{m} = \frac{F}{1} = F$
From equation of motion,
$s = u t + \frac{1}{2} a t^{2} = v + \frac{1}{2} F (1)^{2} = v + \frac{F}{2}$