Particle Motion Problem: Finding Position and Velocity with Derivatives

[chocolatelover]

Homework Statement

A particle moves along the x axis. Its position is given by the equation x=1.5+2.50t-3.9t^2 with x in meters and t in seconds.
a. Determine its position when it changes direction
b. what is its velocity when it returns to the position it had at t=o

Homework Equations

The Attempt at a Solution

a. I would need to take the derivative and set it equal to 0, right?

f'(x)=2.50-7.8t=0
t=(-2.5/-7.8)

I then need to plug t back into the original equation, right?

Would the position when it changes direction be 1.80?

b. What is the velocity when it returns to the position it had at t=0?

2.50-7.8(0)=2.5

Thank you very much

 

[chocolatelover]

Or would this work?

x=1.50-2.50t-3.90t^2

2.5-2(3.90)t

0=2.5-7.8t

t=.321

x=1.5+2.5(.321)-3.9(.321)^2
x=1.9m

vf^2=vi^2+2a(xf-xi)
=2.50^2+2(7.8)(0+1.9)
=6.0

Thank you very much