1. ## Velocity, Accelertion problem

Let s(t)=2t28t+2 for 0t7 denote the position of an object moving along a line. Find

a) Velocity at time t: v(t)=s'(t)= 4t-8

b) Acceleration at time t: v'(t)= 4

c) Initial position: 2(0)-8(0)+2= 2

d) Ending position: 2(7)^2-8(7)+2= 44

--I am having trouble with this part--
e) Total distance traveled: I went the simple route since it started at 2 and ended at 44, I answered 42, but that's not correct.

f) Velocity is positive on the interval: (2,7]

--And this part--
g) Acceleration is positive on the interval [0,??
Since it was a closed interval for the equation, and the acceleration shows to be constant, 4, I tried [0,7], and [0,7), neither of which is right.

2. ## Re: Velocity, Accelertion problem

Originally Posted by JamieG
Let s(t)=2t28t+2 for 0t7 denote the position of an object moving along a line. Find
a) Velocity at time t: v(t)=s'(t)= 4t-8
b) Acceleration at time t: v'(t)= 4
c) Initial position: 2(0)-8(0)+2= 2
d) Ending position: 2(7)^2-8(7)+2= 44
--I am having trouble with this part--
e) Total distance traveled: I went the simple route since it started at 2 and ended at 44, I answered 42, but that's not correct.
f) Velocity is positive on the interval: (2,7]
--And this part--
g) Acceleration is positive on the interval [0,??
Since it was a closed interval for the equation, and the acceleration shows to be constant, 4, I tried [0,7], and [0,7), neither of which is right.
Note that the velocity $v'(t)<0,~0\le t<2$ So $s(0)-s(2)=2-(-6)=8$
Then the particle goes $s(7)-s(2)=~?$