Question

A particle moves on a straight line as such it's product of acceleration and velocity is constant. The distance moved by particle in time t is proportiona​

Answer 1

Answer:

Explanation:

Given, a=k/v

Dv/dt=k/v

Int( Dv.v)=Int(dt)

V^2/2=t+c

V=√2(t+c)

X=int(√2(t+c))

X=2(√2(t+c))^3/2) /3

So distance is directly proportional to t^3/2

This should be the required answer

Thank you!

Answer 2

Answer:

The distance moved by particle in time t is proportional to t^(3/2) .

Explanation:

Given that :

• Product of acceleration and velocity is constant

To find :

• Distance moved by particle in time t

Solution :

• Let, a particle of mass, m having acceleration, a and velocity, v moves on a straight line such that product of acceleration and velocity is constant.
• av = k ---(1)
• where, k is any arbitrary constant.
• Since, av = k.

$a = \frac{k}{v} \\ \frac{dv}{dt} = \frac{k}{v} \\ v \: dv = k \: dt \\ integrating \: above equation, we get \\ ∫v \: dv=∫k \: dt \\ \frac{ {v}^{2} }{2} =kt \\ v= \sqrt{2kt} \\ again, \: putting \: v= \frac{dx}{dt} \: and \: integrating \\ \frac{dx}{dt} = \sqrt{2kt} \\∫ dx = ∫\sqrt{2kt} \: dt \\ x = \frac{2}{3} \sqrt{2k} \: {t}^{ \frac{3}{2} } + c$

• Hence, distance, x moved by particle in time t is proportional to t^(3/2) .