Question

A particle starts from rest and performing circular motion of constant radius with speed given by V= a√x where a is a constant and x is the distance covered. The correct graph of magnitude of its tangential acceleration (at) and centripetal acceleration (ac) versus t will be:​​

Answer 1

Given:

A particle starts from rest and performing circular motion of constant radius with speed given by v= a√x where a is a constant and x is the distance covered.

To find:

Graph between tangential acceleration and centripetal acceleration.

Calculation:

Tangential acceleration can be calculated by first order differentiation of the velocity of the body w.r.t time:

$\therefore \: v = a \sqrt{x}$

$\implies\: a_{t} = \dfrac{dv}{dt}$

$\implies\: a_{t} = \dfrac{d(a \sqrt{x} )}{dt}$

$\implies\: a_{t} = a \times \dfrac{d( \sqrt{x} )}{dt}$

$\implies\: a_{t} = a \times \dfrac{d( \sqrt{x} )}{dx} \times \dfrac{dx}{dt}$

$\implies\: a_{t} = a \times \dfrac{d( \sqrt{x} )}{dx} \times v$

$\implies\: a_{t} = a \times \dfrac{1}{2 \sqrt{x} } \times v$

$\implies\: a_{t} = a \times \dfrac{1}{2 \sqrt{x} } \times a \sqrt{x}$

$\implies\: a_{t} = \dfrac{ {a}^{2} }{2} \: \: \: \: \: \: ......(constant)$

Now, centripetal acceleration:

$\therefore \: a_{c} = \dfrac{ {v}^{2} }{r}$

$\implies \: a_{c} = \dfrac{ {(a \sqrt{x} )}^{2} }{r}$

$\implies \: a_{c} = \dfrac{ {a}^{2} x}{r} \: \: \: \: .....(increasing \: with \: distance)$

So, if we plot a_(t) on Y axis and a_(c) on X axis, then a linear graph parallel to X axis will be obtained.

Refer to attached diagram.