The displacement ' x ' (in meter) 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

Question

The displacement ' x ' (in meter) 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 (A) 6 m (B) 2 m (C) 4 m (D) 0 m

Tardigrade Admin · Accepted Answer

beginarrayl t=√x+3 ⇒ x=(t-3)2 endarray beginarrayl v=(d x/d t)=2(t-3) v=0 Longrightarrow t=3 endarray At t=3, x=(3-3)2=0

Q. The displacement ' $x$ ' (in meter) of a particle of mass ' $m$ ' (in $k g$ ) 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

6 m

2 m

4 m

0 m

$< b r / > < b r / > t = x + 3 < b r / >\Rightarrow x = (t - 3)^{2} < b r / > < b r / >$
$< b r / > < b r / > v = \frac{d x}{d t} = 2 (t - 3) < b r / > v = 0 ⟹ t = 3 < b r / > < b r / >$
At $t = 3, x = (3 - 3)^{2} = 0$

Q. The displacement ' x ' (in meter) 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

6 m

2 m

4 m

0 m

<br/><br/>t=x​+3<br/>⇒x=(t−3)2<br/>​<br/> <br/><br/>v=dtdx​=2(t−3)<br/>v=0⟹t=3<br/>​<br/> At t=3,x=(3−3)2=0

Q. The displacement ' $x$ ' (in meter) of a particle of mass ' $m$ ' (in $k g$ ) 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

$< b r / > < b r / > t = x + 3 < b r / >\Rightarrow x = (t - 3)^{2} < b r / > < b r / >$
$< b r / > < b r / > v = \frac{d x}{d t} = 2 (t - 3) < b r / > v = 0 ⟹ t = 3 < b r / > < b r / >$
At $t = 3, x = (3 - 3)^{2} = 0$