The problem involves a particle moving along a line with a given acceleration function a(t) = t^2 - 7t + 6. 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. We are asked to find the position function s(t).

Here are the steps to solve the problem:

1. Integrate the acceleration function to find the velocity function. The acceleration is the derivative of the velocity, so the velocity is the integral of the acceleration.

∫a(t) dt = ∫(t^2 - 7t + 6) dt = (1/3)t^3 - (7/2)t^2 + 6t + C1

We don't know the constant of integration C1 yet.

2. Use the fact that the velocity at time t=0 is zero to find C1. This comes from the fact that the particle is at rest at t=0.

0 = (1/3)(0)^3 - (7/2)(0)^2 + 6(0) + C1
So, C1 = 0.

Therefore, the velocity function v(t) is v(t) = (1/3)t^3 - (7/2)t^2 + 6t.

3. Integrate the velocity function to find the position function. The velocity is the derivative of the position, so the position is the integral of the velocity.

∫v(t) dt = ∫[(1/3)t^3 - (7/2)t^2 + 6t] dt = (1/12)t^4 - (7/6)t^3 + 3t^2 + C2

We don't know the constant of integration C2 yet.

4. Use the fact that the position at time t=0 is s(0) = 0 to find C2.

0 = (1/12)(0)^4 - (7/6)(0)^3 + 3(0)^2 + C2
So, C2 = 0.

Therefore, the position function s(t) is s(t) = (1/12)t^4 - (7/6)t^3 + 3t^2.

5. However, we have another condition that s(1) = 20. We substitute t=1 into the position function to find if it equals to 20.

20 = (1/12)(1)^4 - (7/6)(1)^3 + 3(1)^2
20 ≠ 20 - 7/6 + 3 = 16 + 1/6

The equation does not hold, which means our position function is incorrect. The discrepancy arises because we have made an assumption that the velocity at time t=0 is zero, which is not given in the problem. Therefore, we cannot solve the problem with the given information. We need more information, such as the velocity of the particle at a certain time, to find the correct position function.

Question

The problem involves a particle moving along a line with a given acceleration function a(t) = t^2 - 7t + 6. 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. We are asked to find the position function s(t).

Here are the steps to solve the problem:

1. Integrate the acceleration function to find the velocity function. The acceleration is the derivative of the velocity, so the velocity is the integral of the acceleration.

∫a(t) dt = ∫(t^2 - 7t + 6) dt = (1/3)t^3 - (7/2)t^2 + 6t + C1

We don't know the constant of integration C1 yet.

2. Use the fact that the velocity at time t=0 is zero to find C1. This comes from the fact that the particle is at rest at t=0.

0 = (1/3)(0)^3 - (7/2)(0)^2 + 6(0) + C1
   So, C1 = 0.

Therefore, the velocity function v(t) is v(t) = (1/3)t^3 - (7/2)t^2 + 6t.

3. Integrate the velocity function to find the position function. The velocity is the derivative of the position, so the position is the integral of the velocity.

∫v(t) dt = ∫[(1/3)t^3 - (7/2)t^2 + 6t] dt = (1/12)t^4 - (7/6)t^3 + 3t^2 + C2

We don't know the constant of integration C2 yet.

4. Use the fact that the position at time t=0 is s(0) = 0 to find C2.

0 = (1/12)(0)^4 - (7/6)(0)^3 + 3(0)^2 + C2
   So, C2 = 0.

Therefore, the position function s(t) is s(t) = (1/12)t^4 - (7/6)t^3 + 3t^2.

5. However, we have another condition that s(1) = 20. We substitute t=1 into the position function to find if it equals to 20.

20 = (1/12)(1)^4 - (7/6)(1)^3 + 3(1)^2
   20 ≠ 20 - 7/6 + 3 = 16 + 1/6

The equation does not hold, which means our position function is incorrect. The discrepancy arises because we have made an assumption that the velocity at time t=0 is zero, which is not given in the problem. Therefore, we cannot solve the problem with the given information. We need more information, such as the velocity of the particle at a certain time, to find the correct position function.

Knowee AI · Accepted Answer