Question

A particle moves along a straight line and its position at time t is given by s(t)=2t3−15t2+24t where s is measured in feet and t in seconds.

a) Find the velocity (in ft/sec) of the particle at time t=0

b) The particle stops moving twice, once when t=A and again when t=B where A 14 years ago

Answer 1

\[v(t)=s'(t)=6t^2-30t+24\]
\[v(0)=24 ft/s\]
b) If it stops moving, the velocity is 0:
\[6t^2-30t+24=0, t^2-5t+4=0, t=1, t=4\]a)\[v(t)=s'(t)=6t^2-30t+24\]
c)\[s(10)=2(10)^3−15(10)^2+24(10)=740 ft \]

Answer 2

how do you figure out the total distance

Answer 3

also can you help me with this

Find equations of the tangent line and normal line to the curve y=10cosx at the point (π/3,5).

Derivative of y= -10sinx

The slope of the tangent line is m1=-10sin(pi/3)
The equation of the tangent line is y=-5sqrt(3)x+(5pi/sqrt3)+5

The slope of the normal line is m2=?
The equation of the normal line is y=?

I just need to know the last two

Answer 4

the particle changes direction at t = 1 and t = 4
displacement at t = 1 = 11 (plug 1 into formula for s(t)
at time = 4 displacement is -16
so it hase travelled 11 + 11 + 16 = 38 up to then
then another 16 + 740 by t=10
total distance travelled = 38 + 16 + 740 = 794 ft

Answer 5

ahh jimmy thanks very much. can you take a look at the question above it too please

Answer 6

not above but below

Answer 7

the one about normal and tangent line