Kinematics Ques 1
1 A particle is moving with speed $v=b \sqrt{x}$ along positive $X$-axis. Calculate the speed of the particle at time $t=\tau$ (assume that the particle is at origin at $t=0$ ).
(2019 Main, 12 April II)
(a) $\frac{b^2 \tau}{4}$
(b) $\frac{b^2 \tau}{2}$
(c) $b^2 \tau$
(d) $\frac{b^2 \tau}{\sqrt{2}}$
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Answer:
Correct Answer: 1.( b )
Solution:
Formula:
Instantaneous Velocity (at an instant) :
- Given, speed, $ v=b \sqrt{x} $
Now, differentiating it with respect to time, we get
$ \frac{d v}{d t}=\frac{d}{d t} b \sqrt{x} $
Now, acceleration
$ \begin{array}{ll} \Rightarrow & a=\frac{b}{2 \sqrt{x}} \cdot \frac{d x}{d t} \\ \Rightarrow & a=\frac{b}{2 \sqrt{x}} \cdot v=\frac{b}{2 \sqrt{x}} \cdot b \sqrt{x}=\frac{b^2}{2} \end{array} $
As acceleration is constant, we use
$ \nu=u+a t $
Now, it is given that $x=0$ at $t=0$.
So, initial speed of particle is
$ u=\left.b \sqrt{x}\right|_{x=0}=b \times 0=0 $
Hence, when time $t=\tau$, speed of the particle using Eq. (i) is
$ v=u+a t=0+\frac{b^2}{2} \cdot \tau=\frac{b^2}{2} \cdot \tau $
BETA
