Question

A particle is moving in a straight line and its distance s, in metres, from a fixed point in the line after t seconds is given by the equation s(t) = t^3 − 12t^2 + 45t. (a) Find the velocity after 2 seconds. (b) Find the acceleration after 2 seconds?

c. after how many seconds is the particle at rest?

Answer 1

`9\ \text{m/s}, \ \ -12\ \text{m/s}^2`

Explanation:

Given the distance `s` of the particle at time `t` is given as

`s=t^3-12t^2+45t`

1) The velocity `v` of the particle is given as follows

`v=\frac{ds}{dt}`

`=\frac{d}{dt}(t^3-12t^2+45t)`

`=3t^2-24t+45`

Now, substituting `t=2` sec, the velocity of particle is

`v=3(2)^2-24(2)+45`

`=9\ \text{m/s}`

2) The acceleration `a` of the particle is given as follows

`a=\frac{dv}{dt}`

`=\frac{d}{dt}(3t^2-24t+45)`

`=6t-24`

Now, substituting `t=2` sec, the velocity of particle is

`v=6(2)-24`

`=-12\ \text{m/s}^2`

Answer 2

`"Velocity after 2 second = 9 m/"s`
`"acceleration after 2 seconds = -12 m/"s^2`

Explanation:

Here ,

`s(t)=t^3 − 12t^2 + 45t. "where , s is in meters ."`

`(I)` Diff. `s(t)` w. r. to `t` , we get Velocity :

`v(t)=s'(t)`

`v(t)=3t^2-24t+45`

Taking `t=2`

`v(2)=3(2)^2-24(2)+45=12-48+45=9` `"m/"s`

Now

`(II)` Diff. `v(t)` w.r.to `t` we get Acceleration :

`a(t)=v'(t)`

`a(t)=6t-24`

Taking `t=2`

`a(2)=6(2)-24=12-24=-12` `"m/"s^2`

Answer 3

(a) `25ms^-1`
(b)`12.5ms^-2`
(c)According to your question, I don't think that the particle will stop at any time as it has a positive acceleration.

Explanation:

(a) `s=2^3-12*2^2+45*2=8-48+90=50` (Assuming that object's distance and displacement are same)
`Velocity=s/t="50m"/"2s"=25ms^-1`

(b)`a=(v-u)/t=(25ms^-1-0ms^-1)/"2s"=12.5ms^-2` (Assuming that object started from stationery position)

(c)According to your question, I don't think that the particle will stop at any time as it has a positive acceleration.