Question

Given the position function, s(t)= t^3/3-15t^2/2 +50t , between t = 0 and t = 12, where s is given in feet and t is measured in seconds, find the interval in seconds where the particle is moving to the left.

Answer 1

\[s(t)= \frac{ t^3 }{ 3 } - \frac{ 15t^2 }{ 2 } + 50t\]

Answer 2

find ds/dt and put it=0
and find t in the given interval

Answer 3

how do I find ds/dt?

Answer 4

is it t^2-15t+50?

Answer 5

@surjithayer ?

Answer 6

t^2-15t+50 <0

Answer 7

5 < t < 10?

Answer 8

correct.

Answer 9

thank you !!

Answer 10

yw