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].

Printable View

- January 12th 2009, 01:01 PMMy Little PonyVelocity, Displacement, Distance
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].

- January 12th 2009, 01:35 PMnvv
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 (Crying), can anyone help me ? - January 12th 2009, 02:20 PMSoroban
Hello, My Little Pony!

Quote:

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.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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.