Question

1) The
position of of a partide varies
with time x= at^2 - bt^3 . The time t
when the acceleraban is equal to zero is​

Answer 1

Answer:

t=a/3b

Explanation:

Answer 2

Given:

The  position of of a particle,x= at²-bt³

To Find:

The time t when acceleration of particle is equal to zero

Solution:

We know that,

• Velocity of a body 'v' is given by

$\pink{\underline{\boxed{\bf{v=\dfrac{dx}{dt}}}}}$

• Acceleration of a body 'a' is given by

$\pink{\underline{\boxed{\bf{a=\dfrac{dv}{dt}}}}}$

where x is displacement or position of particle

━━━━━━━━━━━━━━━━━━━━━━━━━

Let the velocity of particle be v

So,

$\longrightarrow\rm{v=\dfrac{dx}{dt}}$

$\longrightarrow\rm{v=\dfrac{d(at^{2}-bt^{3})}{dt}}$

$\longrightarrow\rm{v=a(2t)-b(3t^{2})}$

$\longrightarrow\rm{v=2at-3bt^{2}}$

Now,

Let the acceleration of particle be 'a'

So,

$\longrightarrow\rm{a=\dfrac{dv}{dt}}$

$\longrightarrow\rm{a=\dfrac{d(2at-3bt^{2})}{dt}}$

$\longrightarrow\rm{a=2a(1)-3b(2t)}$

$\longrightarrow\rm{a=2a-6bt}$

When a=0

$\longrightarrow\rm{0=2a-6bt}$

$\longrightarrow\rm{6bt=2a}$

$\longrightarrow\rm{t=\dfrac{2a}{6b}}$

$\longrightarrow\rm\green{t=\dfrac{a}{3b}}$

Hence, the required time is a/3b.