Question

A particle moves on a straight line and has acceleration a(t)=24t+2. Its position at time t=0 is s(0)=3 and its velocity at time t=0 is v(0)=13. What is its position at time t=5?

Answer 1

It's position at time t = 5 is 593.

Explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

`a(t) = 24t + 2`

Velocity:

`v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^(2) + 2t + K`

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

`v(t) = 12t^(2) + 2t + 13`

Position:

`s(t) = \int {s(t)} \, dt = \int {12t^(2) + 2t + 13} \, dt = 4t^(3) + t^(2) + 13t + K`

Since s(0) = 3

`s(t) = 4t^(3) + t^(2) + 13t + 3`

What is its position at time t=5?

This is s(5).

`s(t) = 4t^(3) + t^(2) + 13t + 3`

`s(5) = 4*5^(3) + 5^(2) + 13*5 + 3`

`s(5) = 593`

It's position at time t = 5 is 593.