A particle moves in the one direction,stops,and then heads back in the opposite direction.The position of a particle (in meters)as a function of time is given by x(t)= -3.65 t² + 4.52 t + 7.91.The position of the particle where it stops is A)0 m. B)9.31 m. C)11.7 m. D)-11.7 m. E)-9.31 m.

Question

Quizplus · Accepted Answer

To find where the particle stops, we need to find where the velocity of the particle is zero, because that is where it changes direction.

Taking the derivative of x(t) with respect to time t gives the velocity function v(t):
v(t) = -3.65 + 4.52 = 0.87

Setting v(t) equal to zero and solving for t gives:
-3.65 t + 4.52 = 0
t = 1.24 s

To find the position where the particle stops, we can plug in t = 1.24 s into the position function x(t):
x(1.24) = -3.65(1.24)^2 + 4.52(1.24) + 7.91
x(1.24) = 9.31 m

Therefore, the answer is B) 9.31 m.

A Particle Moves in the One Direction,stops,and Then Heads Back