A particle is moving with acceleration a(t) = 36 t + 2. its position at time t =0 is s(0) = 16 and its velocity at time t =0 is v(0) = 2.
What is its position at time t = 7
To find the position of the particle at time t = 7, we need to integrate the acceleration function twice to obtain the position function.
Given:
Acceleration a(t) = 36t + 2
Velocity v(t) at time t = 0 is v(0) = 2
Position s(t) at time t = 0 is s(0) = 16
First, integrate the acceleration function to find the velocity function:
∫(a(t)) dt = ∫(36t + 2) dt
v(t) = 18t^2 + 2t + C1
Next, we can use the initial velocity v(0) = 2 to solve for the constant C1:
2 = 18(0)^2 + 2(0) + C1
2 = C1
So, the velocity function becomes:
v(t) = 18t^2 + 2t + 2
Now, integrate the velocity function to find the position function:
∫(v(t)) dt = ∫(18t^2 + 2t + 2) dt
s(t) = 6t^3 + t^2 + 2t + C2
Finally, use the initial position s(0) = 16 to solve for the constant C2:
16 = 6(0)^3 + (0)^2 + 2(0) + C2
16 = C2
Therefore, the position function becomes:
s(t) = 6t^3 + t^2 + 2t + 16
Now, we can substitute t = 7 into the position function to find the position of the particle at time t = 7:
s(7) = 6(7)^3 + (7)^2 + 2(7) + 16
s(7) = 1378
Therefore, the position of the particle at time t = 7 is 1378 units.