Guys as always I appreciate you help. I think this is one of the best sites on the internet. I do however have a question:
A particle moves along a straight line and its position at time t s given by s(t)= 2t^3 - 15 t^2 + 24 t where s is measured in feet and t in seconds. Find the velocity (in ft/sec) of the particle at time t=0___?___. The particle stops moving twice, once when t=A and again when t=B where A<B. A is ___?____
& B is ___?___. What is the position of the particle at time 10 ___?____.
It's velocity is ds/dt = 6 t^2 -30 t +24, so at t=0 ds/dt=24 ft/s.
Originally Posted by qbkr21
The particle stops moving when ds/dt=0, but:
ds/dt = 6 t^2 -30 t +24 = (t-1)(6t-24)
so ds/dt = 0, when t=1 s, and again when t=4 s, so A=1, and B=4.
Now s(t)= 2t^3 - 15 t^2 + 24 t, so at t=10:
s(10) = 2*1000 - 15*100 + 24*10 = 740 ft.