Question

anyone? =(

A particle moves along the x axis. Its position is given by the equation
x = 1.9 + 2.5t − 3.7t2 with x in meters and t in seconds.

(a) Determine its position when it changes direction.
____m

(b) Determine its velocity when it returns to the position it had at t = 0? (Indicate the direction of the velocity with the sign of your answer.)
____m/s

Answer 1

changes direction when the derivative goes from being positive to negative or vice versa

Answer 2

\[x(t)=1.9+2.5t-3.7t^2\]
\[x'(t)=2.5-7.4t\]
put
\[2.5-7.4t=0\] get
\[t=\frac{2.5}{7.4}\]

Answer 3

we could have also done this by simply noting that this as a parabola open down so vertex is at
\[-\frac{b}{2a}\] would have the same answer

Answer 4

but if figured this was cacl class by the next question.

Answer 5

at
\[t=0\] you were at
\[x(0)=1.9\] so set
\[x(t)=1.9\] and solve for t

Answer 6

then substitute that number back in to the derivative to get the velocity.

Answer 7

stupid decimals. i get
\[t=.676\] rounded and
\[x'(.676)=2.4-7.4\times .676\] whatever that is

Answer 8

gonna try it now =)