Question

Help please

The velocity fucntion ( in meters per second) is given for a particle moving along a line . Find the distance traveled by the particle during the given time interval.

Answer 1

\[v(t)= 2t-2 , 0\le t \le5\]

Answer 2

@zepdrix

Answer 3

@myininaya

Answer 4

try integrating the absolute value of the velocity function given on the given interval

Answer 5

okay so itll be
t^2 - 2t ?

Answer 6

|a|=a if a>=0
|a|=-a if a<0

Answer 7

when is v positive and when is v negative on the interval given

Answer 8

okay so what would be my a? would it be the function ?

Answer 9

you are integrating the |v| on [0,5]

Answer 10

a good starting point to figure out when v is positive or negative
is to first find when v is 0

Answer 11

oh okay so then it would be like this
l 2t-2 l ... 2t- 2 >/0

l -2t-2 l ... - 2t-2 < 0

Answer 12

im kind of confuse in this section :/

Answer 13

not exactly

|2t-2| = 2t-2 if 2t-2>=0

|2t-2|=-(2t-2) if 2t-2<0

Answer 14

I just replaced all the a's above with (2t-2)

Answer 15

solve the inequalities

Answer 16

do you know how to solve 2t-2=0?

Answer 17

yes , so itll be t= 1

Answer 18

yes so you will split the integral into two integrals
one for (0,1) and the other (1,5)

Answer 19

we are doing this because v is positive on (1,5)
and v is negative on (0,1)
so |v|=v if t is in (1,5)
and |v|=-v if t is in (0,1)

Answer 20

\[\int\limits_0^1 -(2t-2) dt +\int\limits_1^5 (2t-2) dt\]

Answer 21

oh okay so then take the anti - derivative ?

Answer 22

yep you already did that earlier for v

so you should be at this point:
\[[-(t^2-2t)]_0^1 +[t^2-2t]_1^5\]

Answer 23

oh okay so then the final answer would be 17 ?

Answer 24

17 meters

Answer 25

oh yes

Answer 26

okay question so do we only use the abs values for velocity ?

Answer 27

or when do we use the abs values ?