A particle is moving as given:
v(t)=6sin(t)−3cos(t); s(0)=0
what is the s(t)=?
i tried solving this many times, but i keep getting a wrong answer. thank you in advance for ur help.
remember that v(t) = s'(t)
so if s'(t) = v(t) = 6sin(t)−3cos(t)
then s(t) = -6cost - 3sint + c
given: s(0)=0
0 = -6cos0 - 3sin0 + c
0 = -6 - 0 + c
finish it up
v(t) = ds/dt = 6sin(t)−3cos(t)
integrating ... s(t) = -6 cos(t) - 3 sin(t) + c
for t = 0 ... s(0) = -6 + c ... 0 = -6 + c ... c = 6
s(t) = -6 cos(t) - 3 sin(t) + 6
To find the position function, s(t), we need to integrate the velocity function, v(t), with respect to time.
Given v(t) = 6sin(t) - 3cos(t), we can calculate s(t) using integration.
∫v(t)dt gives us the position function, s(t).
∫(6sin(t) - 3cos(t))dt
We can integrate each term separately since integration is a linear operation.
∫6sin(t)dt - ∫3cos(t)dt
To integrate sin(t), we know that the integral of sin(x) with respect to x is -cos(x).
∫6sin(t)dt = -6cos(t) + C₁
Similarly, the integral of cos(t)dt is sin(t).
∫3cos(t)dt = 3sin(t) + C₂
Note that we introduce constants of integration, C₁ and C₂, since when we differentiate the integral, these constants may emerge.
Therefore, the position function, s(t), is given by:
s(t) = -6cos(t) + C₁ + 3sin(t) + C₂
To determine the values of the constants, we can use the initial condition provided, s(0) = 0.
Plugging in t = 0 in the equation above, we get:
s(0) = -6cos(0) + C₁ + 3sin(0) + C₂
Since sin(0) = 0 and cos(0) = 1, we can simplify the equation to:
0 = -6(1) + C₁ + 3(0) + C₂
0 = -6 + C₁ + C₂
Since the particle starts at s(0) = 0, C₁ + C₂ = 6.
Now, we have a system of equations:
C₁ + C₂ = 6
-6 + C₁ + C₂ = 0
Simplifying the system, we find:
C₁ = 6
C₂ = 0
Now, we can substitute these values back into the position function:
s(t) = -6cos(t) + 6 + 3sin(t) + 0
Simplifying further:
s(t) = -6cos(t) + 3sin(t) + 6
Therefore, the position function, s(t), is -6cos(t) + 3sin(t) + 6.