1) A particle is moving with acceleration a(t)=36t+10. its position at time t=0 is s(0)=16 and its velocity at time t=0 is v(0)=9. What is its position at time t=11?
a(t) = 36t+10
v(t) = 18t^2+10t+c
v(0)=9, so c=9 and
v(t) = 18t^2+10t+9
s(t) = 6t^3+5t^2+9t+c
s(0) = 16, so c=16 and
s(t) = 6t^3+5t^2+9t+16
So, find s(11)
To find the position of the particle at time t=11, we need to integrate the acceleration function twice. The first integration gives us the velocity function, and the second integration gives us the position function.
Given: a(t) = 36t + 10
Integrating this with respect to time (t), we get:
v(t) = ∫(36t + 10) dt
v(t) = 18t^2 + 10t + C1
Using the initial velocity v(0) = 9, we can substitute t=0 and solve for C1:
9 = 18(0)^2 + 10(0) + C1
C1 = 9
Now we have the velocity function:
v(t) = 18t^2 + 10t + 9
Integrating the velocity function with respect to time, we get:
s(t) = ∫(18t^2 + 10t + 9) dt
s(t) = 6t^3 + 5t^2 + 9t + C2
Using the initial position s(0) = 16, we can substitute t=0 and solve for C2:
16 = 6(0)^3 + 5(0)^2 + 9(0) + C2
C2 = 16
Now we have the position function:
s(t) = 6t^3 + 5t^2 + 9t + 16
To find the position at time t=11, substitute t=11 into the position function:
s(11) = 6(11)^3 + 5(11)^2 + 9(11) + 16
s(11) = 798 + 605 + 99 + 16
s(11) = 1518
Therefore, the position of the particle at time t=11 is 1518 units.