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:
$v(t) = \int {a(t)} \, dt = \frac{t^2}{2} + \frac{t^3}{3} + C$
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
$v(t)=3+ \frac{t^2}{2}+ \frac{t^3}{3}$

The position of the particle at time t is the integral of the velocity:
$x(t)=\int {v(t) } \,dt = 3t + \frac{t^3}{6}+ \frac{t^4}{12} +D$
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 
 $x(t)=3t + \frac{t^3}{6}+ \frac{t^4}{12}$ 

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:
$x(5)=3(5) + \frac{5^3}{6}+ \frac{5^4}{12}=87.92$
and the velocity is:
$v(5) = 3+ \frac{5^2}{2}+ \frac{5^3}{3}=57.17$