I asked this before but I don't understand. Could you explain it in easier terms?
The velocity of a subatomic particle moving through space can be modeled by
v(t)=0.1t^2 −2t +0.2
for t≥0 where t is time in seconds and v is velocity in m/s.
Find the following:
a) (a) The time(s) t at which the particle is not moving:
......... and .........
b)The interval(s) over which the particle is moving forward: (Remember to write union as U and infinite as INF) ..................
c)The interval(s) over which the particle is moving backwards: (Remember to write union as U and infinite as INF) ..................
Thanks!
4 answers
I know a is The time(s) t at which the particle is not moving: 0.10 and 19.90
I though B would be [0, 0.10)U(19.90, INF) but it says wrong
I thought c would be -0.10, 19.90 but it's wrong too
I thought c would be -0.10, 19.90 but it's wrong too
Well, if the object is NOT moving, its velocity is obviously zero, so
0.1t^2 −2t +0.2 = 0
t^2 - 20t + 2 = 0
t = (20 ± √392)/2 , clearly we can reject the negative since you said t ≥ 0
t = 10 + √98
0.1t^2 −2t +0.2 = 0
t^2 - 20t + 2 = 0
t = (20 ± √392)/2 , clearly we can reject the negative since you said t ≥ 0
t = 10 + √98
b) moving forward to me means velocity > 0
so t > 10 + √98
c) conversely, to move backwards velocity < 0
which would happen for
0 < t < 10+√98
so t > 10 + √98
c) conversely, to move backwards velocity < 0
which would happen for
0 < t < 10+√98