The velocity function is v(t)= t^2-6t+8 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [0,5].
2007-09-07 12:38:16 · 5 answers · asked by Mr. Rain 1 in Science & Mathematics ➔ Mathematics
At 0 time v(t)= t^2-6t+8, so v(t) = 8
At 5 time v(t)= t^2-6t+8, so v(t) = 25-30+8 = 3
So the velocity at the beginning is 8 and at 5 it is 3
So there has been deceleration of 2 in 5 time units.
2007-09-07 12:41:40 · answer #1 · answered by ignoramus 7 · 0⤊ 0⤋
Find the integral of the function from 0 to 5
the antiderivative (indefinite integral) gives you (t^3)/3 - 3t^2 + 8t + c
Plugging the 5 into the funciton, then subtracting the function with a 0 plugged in gives you:
125/3 - 3(25) + 40 - 0 = 20/3.
The displacement and the distnace traveled by the particle is 20/3 units.
2007-09-07 19:48:02 · answer #2 · answered by ǝɔnɐs ǝɯosǝʍɐ Lazarus'd- DEI 6 · 1⤊ 0⤋
In order to solve this problem and get displacement, you need to integrate the function from 0 to 5.
s = (1/3)t^3 - 3t^2 + 8t + c, this is the integrated function.
Now, substitute 5, and then subtract from this the result of substituting 0.
(125/3 - 75 + 40) - (0)
20/3 is the displacement.
2007-09-07 19:49:42 · answer #3 · answered by Matiego 3 · 0⤊ 0⤋
v(t)=t^2-6t+8
s=integration of v with respect to t from 5 to 0.
s=[t^3/3 -3t^2 + 8t]from 5 to 0
s=5^3/3 - 3(5)^2 +8(5)+ 0^3/3 - 3(0)^2 +8(0)
s=125/3 - 75 + 40
s=6.67m
2007-09-07 19:59:26 · answer #4 · answered by Evans E 1 · 0⤊ 0⤋
what? listen, i'm in middle school...
2007-09-07 19:42:15 · answer #5 · answered by Taom 3 · 0⤊ 1⤋