A particle of mass 40
40kg moves in a straight line such that the force (in newtons) acting on it at time,t (in seconds) is given by, 160t^4-320t^2-360
at time t=0, v is given by v(0)=10, and its position x is given by x(0)=14. What is the position of the particle at time t?
I have that v(t) will be ((4t^5/5)-(8t^3/3)-9t+v0)
Bit stick where to go from here?
Now, use the fact that v(0) = 10
(4t^5/5)-(8t^3/3)-9t+v0 = 10 at t=0
So, v0 = 10, and thus
v(t) = (4t^5/5)-(8t^3/3)-9t+10
Now go on to x(t), using x(0) to find the constant of integration.
To find the position of the particle at time t, given the velocity function v(t), we need to integrate the velocity function with respect to time.
You have correctly calculated the velocity function as follows:
v(t) = (4t^5/5) - (8t^3/3) - 9t + v0
To integrate this function, you can integrate each term separately:
∫(4t^5/5) dt = (4/5) ∫(t^5) dt = (4/5) * (t^(5+1))/(5+1) = (4/5) * (t^6)/6 = (2/15) * (t^6)
∫(-8t^3/3) dt = (-8/3) ∫(t^3) dt = (-8/3) * (t^(3+1))/(3+1) = (-8/3) * (t^4)/4 = (-2/3) * (t^4)
∫(-9t) dt = (-9/1) ∫(t) dt = (-9/1) * (t^(1+1))/(1+1) = (-9/1) * (t^2)/2 = (-9/2) * (t^2)
So, the integrated velocity function becomes:
x(t) = (2/15) * (t^6) + (-2/3) * (t^4) + (-9/2) * (t^2) + c
Where c is the constant of integration.
Given that x(0) = 14, we can substitute t = 0 and x = 14 into the equation to solve for c:
14 = (2/15) * (0^6) + (-2/3) * (0^4) + (-9/2) * (0^2) + c
Simplifying the equation, we find:
14 = c
Therefore, c = 14.
Now we can substitute c back into the integrated velocity function:
x(t) = (2/15) * (t^6) + (-2/3) * (t^4) + (-9/2) * (t^2) + 14
This gives us the position of the particle at time t.