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
The Attempt at a Solution
a. I would need to take the derivative and set it equal to 0, right?
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?
Thank you very much