Question : The displacement ‘x’ (in metre) of a particle of mass ‘m’ (in kg) moving in one dimension under the action of a force, is related to time ‘t’ (in sec) by t=√x+3 The displacement of the particle when its velocity is zero, will be [NEET 2013]
(a) 4 m
(b) 0 m (zero)
(c) 6 m
(d) 2 m
Doubt by Japjee
Solution :
t=√x+3
t-3=√x
Squaring both sides
(t-3)²=(√x)²
t²-6t-9=x
x=t²-6t-9
t-3=√x
Squaring both sides
(t-3)²=(√x)²
t²-6t-9=x
x=t²-6t-9
v=dx/dt
v=d(t²-6t-9)/dt
v=2t-6
Now
v=d(t²-6t-9)/dt
v=2t-6
Now
v=0 (Given)
2t-6=0
2t=6
2t-6=0
2t=6
t=6/2
t=3 sec
Displacement is given by
x=t²-6t-9
Displacement at t=3 sec
=(3)²-6(3)-9
x=t²-6t-9
Displacement at t=3 sec
=(3)²-6(3)-9
=9-18-9
=18-18
=0 m
=18-18
=0 m
Hence, b) Zero, would be the correct option.