First slide

Instantaneous acceleration

Question

A particle moves a distance x in time t according to equation $x = {(t + 5)}^{- 1}$ . The acceleration of particle is proportional to

Moderate

A

(velocity)^3/2
B

(distance)²
C

(distancer)^-2
D

(ve!ocity)^2/3

Solution

Given, $x = \frac{1}{t + 5}$
Differentiating both sides w.r.t. t, we get
   $\frac{d x}{d t} = - \frac{1}{(t + 5)^{2}} \times 1$
$∴$ Velocity, $v = \frac{d x}{d t} = - \frac{1}{(t + 5)^{2}}$ …(i)
Again, differentiating both sides w.r.t. t, we get
   $\frac{d v}{d t} = \frac{1 \times 2}{(t + 5)^{3}} (1) = \frac{2}{(t + 5)^{3}}$ …(ii)
Now, from Eqs. (i) and (ii), we get
   $\frac{d v}{d t} = - 2 v \times \frac{1}{(t + 5)} = (- 2 v) (\sqrt{v}) = - 2 (v)^{3 / 2}$
$∴$ Acceleration, $a = \frac{d v}{d t} = - 2 v^{3 / 2} \Rightarrow a \propto {(velocity)}^{3 / 2}$

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