Question

The acceleration of a particle is given by a = 4t – 30, where a is in meters per second squared and t is in seconds. Determine the velocity and displacement as functions of time. The initial displacement at t = 0 is s0 = -5m, and the initial velocity is vo = 3m/s

Answer 1

The velocity of the particle is given by
`v(t) = 2\cdot t^(2)-30\cdot t + 3`, where
`v` is in meters per second and
`t` is in seconds.

The displacement of the particle is given by
`s(t) = (2)/(3)\cdot t^(3)-15\cdot t^(2)+3\cdot t -5`, where
`s` is in meters and
`t` is in seconds.

Step-by-step explanation:

The acceleration of the particle is given by
`a(t) = 4\cdot t -30`, the expression for velocities are obtained by integration:

Velocity

`v(t) = \int {a(t)} \, dt`

`v(t) = \int {4\cdot t-30} \, dt`

`v(t) = 4\int {t} \, dt -30\int \, dt`

`v(t) = 2\cdot t^(2)-30\cdot t + v_(o)`

Where
`v_(o)` is the initial velocity of the particle, measured in meters per second.

If
`t = 0` and
`v(0) = 3`, then:

`3 = 2\cdot (0)^(2)-30\cdot (0)+v_(o)`

`v_(o) = 3`

The velocity of the particle is given by
`v(t) = 2\cdot t^(2)-30\cdot t + 3`, where
`v` is in meters per second and
`t` is in seconds.

Displacement

`s(t) = \int {v(t)} \, dt`

`s(t) = \int {2\cdot t^(2)-30\cdot t+3} \, dt`

`s(t) = 2\int {t^(2)} \, dt - 30\int {t} \, dt +3\int \, dt`

`s(t) = (2)/(3)\cdot t^(3)-15\cdot t^(2)+3\cdot t +s_(o)`

If
`t = 0` and
`s(0) = -5`, then:

`-5 = (2)/(3)\cdot (0)^(3)-15\cdot (0)^(2)+3\cdot (0)+s_(o)`

`s_(o) = -5`

The displacement of the particle is given by
`s(t) = (2)/(3)\cdot t^(3)-15\cdot t^(2)+3\cdot t -5`, where
`s` is in meters and
`t` is in seconds.