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
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.
2 answers
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
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