A particle is moving with the given data. Find the position of the particle.a(t) = t2 − 9t + 5,    s(0) = 0,    s(1) = 20s(t) =

The problem involves a particle moving along a line with a given acceleration function a(t) = t^2 - 9t + 5. We are also given two initial conditions: the position of the particle at time t=0 is s(0) = 0, and at time t=1 is s(1) = 20.

The position function s(t) can be found by integrating the acceleration function twice.

Step 1: Find the velocity function, v(t), by integrating the acceleration function, a(t):

∫a(t) dt = ∫(t^2 - 9t + 5) dt = (1/3)t^3 - (9/2)t^2 + 5t + C1

This is the velocity function v(t), where C1 is the constant of integration.

Step 2: Use the initial condition s(0) = 0 to find the constant C1.

When t=0, the velocity v(0) is also 0 (since the particle starts from rest). So, we set the velocity function equal to zero and solve for C1:

0 = (1/3)(0)^3 - (9/2)(0)^2 + 5*(0) + C1 => C1 = 0

So, the velocity function is v(t) = (1/3)t^3 - (9/2)t^2 + 5t.

Step 3: Find the position function, s(t), by integrating the velocity function, v(t):

∫v(t) dt = ∫[(1/3)t^3 - (9/2)t^2 + 5t] dt = (1/12)t^4 - (3/2)t^3 + (5/2)t^2 + C2

This is the position function s(t), where C2 is the constant of integration.

Step 4: Use the initial condition s(1) = 20 to find the constant C2.

We set the position function equal to 20 and solve for C2:

20 = (1/12)(1)^4 - (3/2)(1)^3 + (5/2)(1)^2 + C2 => C2 = 20 - (1/12) + 3/2 - 5/2 => C2 = 20 - 1/12 + 2 => C2 = 22 - 1/12 => C2 = 263/12

So, the position function is s(t) = (1/12)t^4 - (3/2)t^3 + (5/2)t^2 + 263/12.