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

 

[Moetasim]

for your help!

Yes, you are on the right track. To find the position when the particle changes direction, you would need to take the derivative of the position equation and set it equal to 0. This will give you the time at which the particle changes direction. Then, you can plug this time back into the original equation to find the position.

For part b, you are correct in finding the velocity when the particle returns to its initial position at t=0. This can be found by plugging in t=0 into the derivative equation. Keep in mind that the units for velocity will be in meters per second.

Overall, your approach is correct and your answers are also correct. Great job!