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A particle starting from rest, moves with a uniform acceleration and covers $x$ meters in the first $5\, second$. The same particle will cover the following distance in the next $5\, seconds$ :-
$x$ metre
$2x$ metre
$3x$ metre
$4x$ metre
Solution
Let the acceleration of the particle be $a$
Initial speed of the particle $u=0 \mathrm{m} / \mathrm{s}$
Distance travelled by it in first 3 seconds, $X=u t+\frac{1}{2} a t^{2} \quad$ where $t=3 \mathrm{s}$ $\therefore X=0 t+\frac{1}{2} a(3)^{2}$
We get $X=4.5 a \quad \ldots .(1)$
Velocity of the particle after $t=3 \mathrm{s}, v=u+a t$
$\therefore v=0+a(3)=3 a$
Distance covered by it next 3 seconds, $Y=v t+\frac{1}{2} a t^{2}$ where $t=3 \mathrm{s}$
$\therefore Y=(3 a) 3+\frac{1}{2} a(3)^{2}=13.5 a$
From (1) and (2), we get $Y=3 X$
