b) find the velocity of each when t = 8
v1 = 2(8)+2 = + 18
v2 = -2t +8 = -8
so in opposite directions
d ) they start out both moving +
when is v1-v2 = 0 ?
two particles are moving along a linear track according to the functions x1=t^2+2t and x2=-t^2+8t+80 for t is greater than or equal to zero.
a)at what value of t will the two particles be in the same place
b)will they be going in the same or opposite directions at the time they are in the same place?
c)at what value of t do they have the same velocity
d)during what time interval(s) are they moving in the same direction?
I need help with parts b and d
a)t^2+2t=-t^2+8t+80
2t^2+2t=8t+80
2t^2-6t-80=0
2(t^2-3t-40)=0
2(t-8)(t+5)=0
t=8 and t=-5, but we reject t=-5 because we cant have negative time
c) v1(t)=2t+2
v2(t)=-2t+8
2t+2=-2t+8
4t+2=8
4t=6
t=3/2
3 answers
they both start out +
find out when the first one hits v = 0
then it goes - while the first is +
find when the second hits v=0
then they both are -
etc
find out when the first one hits v = 0
then it goes - while the first is +
find when the second hits v=0
then they both are -
etc
oh, the first one is always v positive v = 2t+2
so when does the second one go - ?
so when does the second one go - ?