A particle is moving with the velocity v(t), given by the function below.Find the total displacement of the particle in the first 10 seconds.

Question

v(t)=cos(πt/6)+sin(πt/9)

Answer 1

By definition, displacement is the difference between the final position and the initial position. We can symbolize this as follows:

displacement=final position-initial position=x(10)-x(0), where x(t) is the position of the particle at time t, x(0) is the initial and x(10) is the final position

We are given the velocity function v(t)=cos(πt/6)+sin(πt/9), which we can integrate to obtain the position function. This yields x(t)=6/πsin(πt/6)-9/πcos(πt/9)+c. Substituting this function into the displacement expression above we obtain:

displacement=x(10)-x(0)=[6/πsin(π10/6)-9/πcos(π10/9)+c]-[6/πsin(0)-9/πcos(0)+c]=

6/πsin(π10/6)-9/πcos(π10/9)+9/π

Answer 2

I'll use upper case S as the integral sign and lower case s to represent position

We know Sv(t) dt from a to b= s(b)-s(a)

We can also assume at t = 0, the particle started at 0 displacement so s(0) = 0

Sv(t)dt = S[cos(πt/6)+sin(πt/9)]dt from t = 0 to t = 10

You can use substitution for this integral.

Sv(t)dt = (6/π)sin(πt/6)-(9/π)cos(πt/9) from t = 0 to t = 10

Remember Sv(t) dt from a to b = s(b)-s(a)

s(10)-s(0)=(6/π)sin(πt/6)-(9/π)cos(πt/9) from t = 0 to t = 10

s(10) - 0 =[(6/π)sin(π10/6)-(9/π)cos(π10/9)] - [(6/π)sin(π0/6)-(9/π)cos(π0/9)]

s(10) = [(6/π)sin(π5/3)-(9/π)cos(π10/9)] - [(6/π)sin(0)-(9/π)cos(0)]

s(10) = [(6/π)sin(π10/6)-(9/π)cos(π10/9)] - [-9/π]

s(10) = [(6/π)sin(π10/6)-(9/π)cos(π10/9)] + 9/π

Using a calculator

s(10)=3.902 in what ever unit of distance is being used here

Answer 3

Distance D(t) = Integral V(t) = (6/π)Sin(πt/6) - (9/π)Cos(πt/9)

D(0) = -2.865

D(10) = 1.038

Displacement is 1.038 - (-2.865) = 3.903