Question 89: Motion in 1D

Question

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity
'v' that varies as $$v = \alpha \sqrt x $$. The displacement of the particle varies with time as

Accepted Answer

Given the velocity function: $$v = \alpha \sqrt{x}$$
We know that velocity $v$ is the rate of change of displacement with respect to time:
$$	herefore \frac{dx}{dt} = \alpha \sqrt{x}$$
Rearranging and separating variables, we get:
$$\Rightarrow \frac{dx}{\sqrt{x}} = \alpha \, dt$$
To solve this, we integrate both sides:
$$\int\limits_{0}^{x} \frac{dx}{\sqrt{x}} = \alpha \int\limits_{0}^{t} dt$$
This gives us:
$$\Rightarrow \left[ \frac{2\sqrt{x}}{1} ight]_{0}^{x} = \alpha \left[tight]_{0}^{t}$$
Simplifying, we obtain:
$$\Rightarrow 2\sqrt{x} = \alpha t $$
Squaring both sides:
$$\Rightarrow x = \frac{\alpha^2}{4} t^2$$
Thus, the displacement $x$ varies with time $t$ as:
$$x \propto t^2$$

Answer

t2

Answer

t

Answer

t1/2

Answer

t3

Motion in 1D

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity 'v' that varies as $v = \alpha \sqrt x $. The displacement of the particle varies with time as

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