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
a = 36 t + 2
v = 18 t^2 + 2 t + Vo but Vo = 2
s = 6 t^3 + t^2 + Vi t + So but So = 16
so
s = 6 t^3 + t^2 + 2 t + 16
put in 7 for t
To find the position of the particle at time t = 7, we need to integrate the given acceleration function twice.
First, let's find the velocity function by integrating the acceleration function with respect to time.
∫(a(t)) dt = ∫(36t + 2) dt
= 18t^2 + 2t + C1
Since we know the initial velocity v(0) = 2, we can substitute this value into the equation to solve for the constant C1:
v(0) = 18(0)^2 + 2(0) + C1
= 0 + 0 + C1
= C1
So, C1 = 2.
Now we have the velocity function v(t) = 18t^2 + 2t + 2.
Next, we need to find the position function by integrating the velocity function with respect to time.
∫(v(t)) dt = ∫(18t^2 + 2t + 2) dt
= 6t^3 + t^2 + 2t + C2
Since we know the initial position s(0) = 16, we can substitute this value into the equation to solve for the constant C2:
s(0) = 6(0)^3 + (0)^2 + 2(0) + C2
= 0 + 0 + 0 + C2
= C2
So, C2 = 16.
Therefore, the position function is s(t) = 6t^3 + t^2 + 2t + 16.
To find the position at time t = 7, substitute t = 7 into the position function:
s(7) = 6(7)^3 + (7)^2 + 2(7) + 16
= 1176 + 49 + 14 + 16
= 1255
Therefore, the position of the particle at time t = 7 is 1255.