You are watching an object that is moving in shm. when the object is displaced 0.650 m to the right of its equilibrium position, it has a velocity of 2.50 m/s to the right and an acceleration of 8.40 m/s2 to the left. how much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

Explanation:

Let's model with sine:

x(t) = Asin(ωt)

so that t = 0, x = 0

x(t) = 0.65 m = A*sin(ω*t)

v(t) = x(t)'= 2.50 m/s = A*ω*cos(ω*t)

a(t) = v(t)'= -8.40 m/s² = -A*ω²*sin(ω*t)

x(t) / a(t) = Asin(ωt) / -Aω²sin(ωt)

0.65m / -8.40 m/s² = -1 / ω²

ω² = 12.934 rad^2/s^2

ω = 3.59 rad/s  

x(t) / v(t) = Asin(ωt) / Aωcos(ωt)

0.650m / 2.50m/s = tan(3.59t) / 3.59 rad/s

0.9334 = tan(3.59t)

t = 0.176 s  

x(0.176) = A*sin(3.59*0.176)

0.65 m= A*sin(0.631)

A = 0.732 m ← amplitude of motion

your answer is yx

hope this

~mia

car a and b are moving in opposite directions

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