Question

the distance in kilometres of a streetcar away from the station is modellled by d(t)= - 6 sin (pi/10(t+5))+6 where t is the time in minutes. each round trip take 20 minutes. When will the streetcar be 8km away from the station in the next 30 mins

Answer 1

$- 6 sin (π/10(t+5))+6=8 \\ - 6 sin (π/10(t+5))=8-6 \\ - 6 sin (π/10(t+5))=2 \\ sin (π/10(t+5))=-2/6 \\ sin (π/10(t+5))=-1/3 \\ π/10(t+5))=(-1)^k arcsin(-1/3) +π*k, \ k∈Z \\ t+5=10*((-1)^k arcsin(-1/3) +π*k)/π, \ k∈Z \\ t=10*((-1)^k arcsin(-1/3) +π*k)/π -5, \ k∈Z \\ t=10*((-1)^{k+1} arcsin(1/3) +π*k)/π -5, \ k∈Z \\ t=\mp10arcsin(1/3)/π -5+20k, \ k∈Z \\ t\approx\mp6.08173+20k, \ k∈Z \\ k=0 ,\ t\approx6.1 \ (min) \\ k=1 ,\ t\approx 13.9 \ (min) \\ k=1 ,\ t\approx26.1\ (min) \\ Answer: \ 6.1 min,\ 13.9 min, \ 26.1 min$