Displacement x of a particle moving in one dimension is related to time t by the equation \[t=\sqrt{x}+2\]. The displacement of the particle when its velocity is zero, is (Here x is in metre and t is in second)
A). 4
B). 2
C). 1
D). 0
Answer
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Hint: Velocity of the particle is given by rate of change of displacement of the particle. It is the differentiation of displacement with respect to time t. It's given that velocity is zero so it should be remembered that a differentiation of a constant is zero.
Complete step by step answer:
The velocity of a particle is given by rate of change of displacement per unit time: $x=0$
$v=\dfrac{dx}{dt}$
Now, $t=\sqrt{x}+2$
Hence, $x={{(t-2)}^{2}}$
On substituting this in above equation we get:
$v=\dfrac{d{{(t-2)}^{2}}}{dt}$
$v=\dfrac{d({{t}^{2}}+4-4t)}{dt}$
On differentiating we get;
$v=2t-4$
Since v=0
We get
$t=2$
Substituting this value of t in the equation given in the question, we get:
$2=\sqrt{x}+2$
Therefore,
$x=0$
Hence, the correct answer is option D. 0
Note: Students must remember that a particle moving in a straight line is stationary when its velocity is zero as it is obtained in the above solution. The velocity of the particle is a derivative of position and the acceleration is a derivative of velocity.
Complete step by step answer:
The velocity of a particle is given by rate of change of displacement per unit time: $x=0$
$v=\dfrac{dx}{dt}$
Now, $t=\sqrt{x}+2$
Hence, $x={{(t-2)}^{2}}$
On substituting this in above equation we get:
$v=\dfrac{d{{(t-2)}^{2}}}{dt}$
$v=\dfrac{d({{t}^{2}}+4-4t)}{dt}$
On differentiating we get;
$v=2t-4$
Since v=0
We get
$t=2$
Substituting this value of t in the equation given in the question, we get:
$2=\sqrt{x}+2$
Therefore,
$x=0$
Hence, the correct answer is option D. 0
Note: Students must remember that a particle moving in a straight line is stationary when its velocity is zero as it is obtained in the above solution. The velocity of the particle is a derivative of position and the acceleration is a derivative of velocity.
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