Question #be972

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Answer 1 · 2016-10-17T10:06:23

Let the distance between p and q be S m and the accelaration of the particle is $" "a" m/s^2$ . Let the total time to cover distance S is T s.

The particle starts with initial velocity $u=0$ and describes 2 m in 1st sec.

So applying kinematic equation
$s=ut+1/2at^2$
$2=0xx1+1/2xxaxx1^2$
$=>a=4m/s^2$

Again $S=0xxT+1/2xx4xxT^2$

$=>S=2T^2$

Distance covered in $(T-3)$ s

$S_((T-3))=0xx(T-3)+1/2xx4(T-3)^2=2(T-3)^2$

Si the distance covered in last 3 sec

$S-S_((T-3))=11/16S=2T^2-2(T-3)^2$

$=>11/16xx2T^2=2T^2-2(T-3)^2$

$=>(T-3)^2=5/16T^2$

$=>(T-3)=sqrt5/4T$

$=>(4-sqrt5)/4T=3$

$=>T=12/(4-sqrt5)s=12/11(4+sqrt5)s$

So the dustance between p and q

$S=2T^2=12^2/11^2(4+sqrt5)^2m=46.28m$