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Question

Physics Question on distance and displacement

The displacement x'x' (in meter) of a particle of mass m'm' (in kgkg) moving in one dimension under the action of a force, is related to time t't' (in sec) by t=x+3t = \sqrt x +3. The displacement of the particle when its velocity is zero, will be

A

4 m

B

0 m (zero)

C

6 m

D

2 m

Answer

0 m (zero)

Explanation

Solution

Given : t=x+3t=\sqrt{x}+3
or x=t3\sqrt{x}=t-3
Squaring both sides, we get
x=(t3)2x=(t-3)^{2}
Velocity, v=dxdt=ddt(t3)2=2(t3)v=\frac{d x}{d t}=\frac{d}{d t}(t-3)^{2}=2(t-3)
Velocity of the particle becomes zero, when
2(t3)=02(t-3)=0
or t=3st=3 s
At t=3st=3 s
x=(33)2=0mx=(3-3)^{2}=0\, m