Question

A particle moves with acceleration function a(t) = 5 + 4t - 2t^2. Its initial velocity v(0) = 3 m/s and its initial displacement is s(0) = 10 m. Find its position after 4 seconds.

Answer 1

Its position after 4 seconds is 62 meters.

Step-by-step explanation:

It is given that,

The acceleration of the particle is given by equation :

`a(t)=5+4t-2t^2`

Also,
`a=(dv)/(dt)`

`v=\int\limits {a.dt}`

`v=\int\limits {(5+4t-2t^2).dt}`

`v=5t+2t^2-(2)/(3)t^3+c`

At t = 0,
`v(0)=3\ m/s`. So, c = 3

`v=5t+2t^2-(2)/(3)t^3+3`

Also,
`v=(ds)/(dt)`, s is the position

`s=\int\limits {v.dt}`

`s=\int\limits {(5t+2t^2-(2)/(3)t^3+3).dt}`

`s=(5)/(2)t^2+(2)/(3)t^3-(t^4)/(6)+3t+c'`

At t = 0,
`s(0)=10\ m`. So, c' = 10

`s=(5)/(2)t^2+(2)/(3)t^3-(t^4)/(6)+3t+10`

At t = 4 s

`s=(5)/(2)(4)^2+(2)/(3)(4)^3-((4)^4)/(6)+3(4)+10`

s = 62 m

So, at t = 4 seconds the position of the particle is 62 meters. Hence, this is the required solution.