Question

A particle travels along the x-axis in such a way that its acceleration at time t is a(t) = t + t2. if it starts at the origin with an initial velocity of 3 (that is, s(0) = 0 and v(0) = 3), determine its position and velocity when t = 5. (round your answers to two decimal places.)

Answer 1

The acceleration of the particle as a function of time t is

`a(t) = t + t^2`
The velocity of the particle at time t is the integral of the acceleration:

`v(t) = \int {a(t)} \, dt = (t^2)/(2) + (t^3)/(3) + C`
where the constant C can be found by requiring that the velocity at time t=0 is v=3:

`v(0) = 3`
and we find
`C=v_0=3`
so the velocity is

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

The position of the particle at time t is the integral of the velocity:

`x(t)=\int {v(t) } \,dt = 3t + (t^3)/(6)+ (t^4)/(12) +D`
where D can be found by requiring that the initial position at time t=0 is zero:
x(0)=0
from which we find D=0, so

`x(t)=3t + (t^3)/(6)+ (t^4)/(12)`

To solve the problem, now we just have to substitute t=5 into x(t) and v(t) to find the position and the velocity of the particle at t=5.

The position is:

`x(5)=3(5) + (5^3)/(6)+ (5^4)/(12)=87.92`
and the velocity is:

`v(5) = 3+ (5^2)/(2)+ (5^3)/(3)=57.17`