Question

3. 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 xit)s -2.68t+252 t+3.91. What is the position of the particle where it stops?
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Answer #1

"When it stops" means "When its velocity is zero"

x(t)= -2.68t² + 2.52t + 3.91

velocity = dx/dt = -5.36t + 2.52
When dx/dt = 0
-5.36t + 2.52 = 0
t = 2.52/5.36 = 0.470 s

So what is x(t) when t = 0.470 ? Plug that value for t back into the original expression:

x(t)= -2.68 * 0.470² + 2.52 * 0.470 + 3.91
x = -0.5920 + 2.99 + 3.91
x = 6.308 meters

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