Question

A particle moves in a straight line, so that x= √t. Then it's acceleration is proportional to? (1)velocity (2) (velocity)^3/2 (3) ( velocity)^3. (4) ( velocity)^2

please answer fast​

Answer 1

dx\dt=d\dt(t)½

V=1\2t-½

t=√1\2v

A=V\T

A=V\√2v

A=2V³\²

Answer is velocity^3\2

Answer 2

Answer:

Option(3)

Explanation:

We have,

$x=\sqrt{t}$

Differentiating once gives velocity,

$v= \frac{dx}{dt}\\\\\frac{x}{y} = \frac{d \sqrt{t} }{dt} \\\\= \frac{d(t^{\frac{1}{2} }) }{dt }\\\\=\frac{1}{2} t^{-\frac{1}{2} }$-----(i)

Differentiating it once again gives acceleration,

$a = \frac{dv}{dt} \\\\= \frac{d(\frac{1}{2}t^{-\frac{1}{2} } )}{dt} \\= \frac{1}{2}(-\frac{3}{2} ) t^{-\frac{3}{2} } \\\\= -\frac{3}{4} \frac{1}{\sqrt{t^{3} } }$

Since we already had an equation for velocity therefore by using (i)

$\frac{-3}{2} (t^{\frac{-1}{2}})^{3} \\=\frac{-3}{2} (2v)^{3}$

From here he have,

acceleration directly proportional to $(velocity)^{3}$.

Hence option (3) is the correct one.