Question

S=t^3-6t^2+9t, find the velocity at 2 sec and acceleration at t=3 sec

Answer 1

Answer:

s=t^3-6t^2+9t

ds/dt=d(t^3-6t^2+9t)/dt. {ds/dt=v}

v=d(t^3-6t^2+9t)/dt

v=3t^2-6t+9

so,at t=2sec

v=3×2^2 -6×2+9

v=12-12+9

v=9m/s

now, from above

v=3t^2-6t+9

now differentiate v with respect to t

dv/dt=d(3t^2-6t+9)/dt. {dv/dt=a}

a=d(3t^2-6t+9)/dt.

a=6t-6

so,at t=3sec

a=(6×3)-6

a=18-6

a=12m/s^2

Answer 2

Given:

Position of a body is given by

• s (t) = t³ - 6 t² + 9 t

To find:

• velocity of body at 2 sec, v(2) =?
• Acceleration of body at 3 sec, a(3) = ?

Knowledge required:

• Velocity 'v' is given by the differential coefficient of displacement 's' with respect to time 't'.

$\;\;\;\;\;\;\;\boxed{\sf{v=\dfrac{d\;s}{d\;t}}}$

• Acceleration 'a' is given by the differential coefficient of velocity ;v; with respect to time 't'.

$\boxed{\sf{a=\dfrac{d\;v}{d\;t}}}$

Solution:

Using formula

$\implies\sf{v=\dfrac{d\;s}{d\;t}}$

$\implies\sf{v(t)=\dfrac{d\;\;(t^3-6t^2+9t)}{d\;t}}$

differentiating in RHS we will get

$\implies\sf{v(t)=3t^2-12t+9}$

We need to calculate velocity of body at 2 sec therefore,

$\implies\sf{v(2)=3\times(2)^2-12\times(2)+9}$

$\implies\boxed{\boxed{\sf{v(2)=12-24+9=-3\;ms^{-1}}}}$

Now,

Using formula

$\implies\sf{a=\dfrac{d\;v}{d\;t}}$

$\implies\sf{a(t)=\dfrac{d\;\;(3t^2-12t+9)}{d\;t}}$

differentiating in RHS we will get

$\implies\sf{a(t)=6t-12}$

We need to find acceleration at time 3 sec therefore,

$\implies\sf{a(3)=6(3)-12}$

$\implies\boxed{\boxed{\sf{a(3)=18-12=6\;ms^{-2}}}}$

Therefore,

• velocity of body at time 2 sec is  -3 ms⁻¹.
• Acceleration of body at time 3 sec is 6 ms⁻².