The velocity function is v(t)= -t^2 + 6t -8 for a particle moving along a line. Find the displacement and the distance travelled by the particle during the time interval [-1, 5].
Hi, My Little Pony
displacement = integral from -1 to 5 (-t^2 + 6t -8) = -18
distance = Sqrt[ ((-5^2+6*5-8) - (1^2-6*1-8))^2) ] = Sqrt[((-3)-(-12))^2]=Sqrt[12^2] = 12 (I'm not sure about this but it looks correctly)
p.s. I'm new on this forum and I don't now how to use math symbols , can anyone help me ?
Hello, My Little Pony!
We have: .The velocity function is: for a particle moving along a line.
Find the displacement and the distance travelled on the time interval [-1, 5].
The position function is: .
Assume that , then:
The displacement is: .
The particle is displaced 18 units to the left.
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Solve
. . . This is where the particle stops ... and changes directions.
In the first two seconds, its displacement is: .
It has moved units to the left.
In the next two seconds, its displacement is: .
It has moved units to the right.
In the last second, its displacement is: .
It has moved units to the left.
It has moved a total distance of: . units.