Let a(t), v(t) and s(t) be the acceleration, velocity, and position functions respectively of a particle. If a(t)=5cos(t/2)+sint and s(0)=0, v(0)=-5. How do i find the position function s(t)??
find s(t) as the second integral of a(t)
s(t)= INTINT a(t) dt dt
= INT (10sint/2 -cost + c)dt
= - 20cost/2 - sint + ct + k where c, k are consttants.
Now evaluate v(0) to find c, and s(o) to find c.
To find the position function s(t), you need to integrate the acceleration function a(t) twice with respect to time.
First, integrate a(t) with respect to t to find the velocity function v(t):
∫ a(t) dt = ∫ (5cos(t/2) + sin(t)) dt
Integrating each term separately:
= ∫ 5cos(t/2) dt + ∫ sin(t) dt
Using the integration rules:
= 10sin(t/2) - cos(t) + C1
where C1 is the constant of integration.
Next, integrate v(t) with respect to t to find the position function s(t):
∫ v(t) dt = ∫ (10sin(t/2) - cos(t) + C1) dt
Integrating each term separately:
= ∫ 10sin(t/2) dt - ∫ cos(t) dt + ∫ C1 dt
Using the integration rules:
= -20cos(t/2) - sin(t) + C1t + C2
where C2 is the constant of integration.
Therefore, the position function s(t) is:
s(t) = -20cos(t/2) - sin(t) + C1t + C2
To find the values of the constants C1 and C2, you can use the initial conditions given in the problem:
s(0) = 0: Substitute t = 0 into the position function and set it equal to 0:
0 = -20cos(0/2) - sin(0) + C1(0) + C2
0 = -20(1) - 0 + 0 + C2
C2 = 20
v(0) = -5: Substitute t = 0 into the velocity function and set it equal to -5:
-5 = 10sin(0/2) - cos(0) + C1
-5 = 10(0) - 1 + C1
C1 = -4
Substituting the values of C1 and C2 back into the position function gives the final position function:
s(t) = -20cos(t/2) - sin(t) - 4t + 20