A particle is moving along the x-axis so that its position at any time [tex]t \geq 0[/tex] is given by [tex]x(t) = 2te^{-t}[/tex].

A) Find [tex]v(t)[/tex].
B) Find [tex]a(t)[/tex].
C) At [tex]t = 0[/tex], is the particle slowing down or speeding up? Explain.
D) At what time(s), if any, does the particle change directions? Justify your answer.

for speed you can differentiate the equation, for acceleration you can again differentiate the equation . at t=0 the particle is slowing down , when you get equation for velocity put t=0 then only -1 is left

Answered by Shashank | 2024-06-10

The velocity function is v ( t ) = 2 e − t ( 1 − t ) and the acceleration function is a ( t ) = − 2 e − t ( 2 − t ) . At t = 0 , the particle is slowing down since it has positive velocity and negative acceleration. The particle changes direction at t = 1 when the velocity becomes zero, confirming it changes from moving forward to moving backward.
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Answered by Shashank | 2024-12-23