There are two particles (1 and 2) that are moving around in space. The force that particle 2 exerts on 1 is given by:

F21(t)=Fx*e^−(t/T)i^ + Fy*sin(2πt/T) j^

Where the parameters have the values:
Fx = 13.4 N
Fy = -84.2 N
T = 74 s
tf = 140.46 s

Find the average force from 1 on 2 from t = 0 until t = tf

Find the average force from 2 on 1 from t = 0 until t = tf

To find the average force from particle 1 on particle 2, we need to integrate the force function F21(t) over the time interval from t = 0 to t = tf and then divide by the total time interval (tf - 0).

1. Integrating F21(t) over the interval [0, tf]:
∫[0,tf] F21(t) dt = ∫[0,tf] (Fx*e^(-t/T)i^ + Fy*sin(2πt/T) j^) dt

The integral of Fx*e^(-t/T)i^ with respect to t is:
∫(Fx*e^(-t/T)i^) dt = -Fx*T*e^(-t/T)i^

The integral of Fy*sin(2πt/T) j^ with respect to t is:
∫(Fy*sin(2πt/T) j^) dt = -Fy*T/(2π) * cos(2πt/T) j^

So, the integral of F21(t) over the interval [0, tf] is:
∫[0,tf] F21(t) dt = -Fx*T*e^(-t/T)i^ - Fy*T/(2π) * cos(2πt/T) j^

2. Calculating the average force from particle 1 on particle 2:
AverageForce_1on2 = (1/(tf - 0)) * ∫[0,tf] F21(t) dt

Substituting the integral we calculated earlier:
AverageForce_1on2 = (1/(tf - 0)) * (-Fx*T*[e^(-t/T)]|[0,tf])i^ - (1/(tf - 0)) * (Fy*T/(2π) * cos(2πt/T) | [0, tf])j^

Simplifying:
AverageForce_1on2 = (-Fx*T*e^(-tf/T) + Fx*T)i^ - (Fy*T/(2π) * (cos(2πtf/T) - 1))j^

Now, let's find the average force from particle 2 on particle 1 using a similar process.

To find the average force from particle 2 on particle 1, we need to integrate the force function F12(t) over the time interval from t = 0 to t = tf and then divide by the total time interval (tf - 0).

The force function F12(t) is the negative of F21(t) since it represents the force exerted by particle 2 on particle 1.

So, F12(t)= -Fx*e^(-t/T)i^ - Fy*sin(2πt/T) j^

Therefore, the average force from particle 2 on particle 1 is:

AverageForce_2on1 = (1/(tf - 0)) * ∫[0,tf] F12(t) dt

Substituting in the force function F12(t) and following the same process as above, we can calculate the average force from particle 2 on particle 1.

Please note that the values of Fx, Fy, T, and tf provided in the question need to be substituted into the equations to obtain the final numerical values.

To find the average force from particle 1 on particle 2 from t = 0 until t = tf, we need to integrate the force function over that time interval and divide by the total time.

The force function F21(t) is given as:
F21(t) = Fx * e^(-t/T)i^ + Fy * sin(2πt/T)j^

We need to integrate this force function with respect to time from 0 to tf.

Let's start by integrating the x-component of the force, Fx * e^(-t/T), with respect to time:
∫(0 to tf) Fx * e^(-t/T) dt

To solve this integral, we can use the rule ∫e^x dx = e^x + C:
∫(0 to tf) Fx * e^(-t/T) dt = Fx * ∫(0 to tf) e^(-t/T) dt

Now we integrate e^(-t/T):
Let z = -t/T
dz = -dt/T

When t = 0, z = 0
When t = tf, z = -tf/T

Now the integral becomes:
Fx * ∫(0 to -tf/T) e^z (-T dz)
= -Fx * T * ∫(0 to -tf/T) e^z dz
= -Fx * T * [-e^z] (0 to -tf/T)
= -Fx * T * [-e^(-tf/T) + e^0] = Fx * T * (1 - e^(-tf/T))

Now let's move on to integrating the y-component of the force, Fy * sin(2πt/T):
∫(0 to tf) Fy * sin(2πt/T) dt

To solve this integral, we can use the rule ∫sin(ax) dx = -1/a * cos(ax) + C:
∫(0 to tf) Fy * sin(2πt/T) dt = Fy * (T/2π) * [-cos(2πt/T)] (0 to tf)
= Fy * (T/2π) * [-cos(2π(tf)/T) + 1]

Now we can find the average force from particle 1 on particle 2 by dividing the total force by the total time:
Average Force from 1 on 2 = (∫(0 to tf) F21(t) dt) / tf
= [(Fx * T * (1 - e^(-tf/T)))i^ + (Fy * (T/2π) * [-cos(2π(tf)/T) + 1])j^] / tf

Similarly, to find the average force from particle 2 on particle 1, we integrate the force function F12(t) = -F21(t) from t = 0 to tf.
Thus, the integration for the average force from 2 on 1 would give:
Average Force from 2 on 1 = (∫(0 to tf) -F21(t) dt) / tf
= [(-Fx * T * (1 - e^(-tf/T)))i^ + (-Fy * (T/2π) * [-cos(2π(tf)/T) + 1])j^] / tf

Now you can substitute the given values for Fx, Fy, T, and tf to calculate the average forces.