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!
a) v(t)=0.1t^2 −2t +0.2 for t>=0
v = 0
t^2 - 20 t + 2 = 0
t = [ 20 +/- sqrt( 400 - 8 ) ] / 2
t = 10 +/- 0.5(sqrt 392)
t = 10 +/- 9.90
t = 20 or 0.1
it is negative between those points, when t -->inf +0.1 t^2 wins
a) The particle is not moving when its velocity is equal to zero. So, we can solve the equation v(t) = 0:
0.1t^2 - 2t + 0.2 = 0
Using the quadratic formula, we can find the values of t where the particle is not moving:
t = (-(-2) ± √((-2)^2 - 4(0.1)(0.2))) / (2(0.1))
Simplifying it:
t = (2 ± √(4 - 0.08)) / 0.2
t = (2 ± √3.92) / 0.2
t ≈ (2 ± 1.98) / 0.2
t ≈ (3.98 / 0.2) or (0.02 / 0.2)
t ≈ 19.9 or 0.1
So, the particle is not moving at t=19.9 seconds and t=0.1 seconds.
b) To determine when the particle is moving forward, we need to find the intervals where its velocity is positive. So, we need to solve the inequality v(t) > 0:
0.1t^2 - 2t + 0.2 > 0
We can solve this inequality by factoring or using the quadratic formula. Factoring is difficult, so let's use the quadratic formula:
t = (-(-2) ± √((-2)^2 - 4(0.1)(0.2))) / (2(0.1))
Simplifying it:
t = (2 ± √(4 - 0.08)) / 0.2
t = (2 ± √3.92) / 0.2
t ≈ (2 ± 1.98) / 0.2
t ≈ (3.98 / 0.2) or (0.02 / 0.2)
t ≈ 19.9 or 0.1
So, the particle is moving forward for t>0.1 seconds and t<19.9 seconds.
c) To determine when the particle is moving backwards, we need to find the intervals where its velocity is negative. So, we need to solve the inequality v(t) < 0:
0.1t^2 - 2t + 0.2 < 0
Again, let's use the quadratic formula:
t = (-(-2) ± √((-2)^2 - 4(0.1)(0.2))) / (2(0.1))
Simplifying it:
t = (2 ± √(4 - 0.08)) / 0.2
t = (2 ± √3.92) / 0.2
t ≈ (2 ± 1.98) / 0.2
t ≈ (3.98 / 0.2) or (0.02 / 0.2)
t ≈ 19.9 or 0.1
Since the inequality is strict, the particle is not moving backward at any time.
Therefore:
a) The particle is not moving at t = 19.9 seconds and t = 0.1 seconds.
b) The particle is moving forward for t > 0.1 seconds and t < 19.9 seconds.
c) The particle is not moving backward at any time.
To find the time(s) at which the particle is not moving, we need to find the t-values where the velocity is zero.
a) Setting v(t) = 0, we have:
0.1t^2 - 2t + 0.2 = 0
Solving this quadratic equation, we can either factor it or use the quadratic formula.
Factoring:
0.1t^2 - 2t + 0.2 = 0
t(0.1t - 2) + 0.2 = 0
(0.1t - 2)(t + 0.1) = 0
Setting each factor to zero:
0.1t - 2 = 0 or t + 0.1 = 0
Solving each equation:
0.1t = 2 or t = -0.1
t = 20 or t = -0.1
Since time cannot be negative, we disregard t = -0.1.
Therefore, the time(s) at which the particle is not moving is t = 20.
b) To find the interval(s) over which the particle is moving forward, we need to consider when the velocity is positive.
Since v(t) = 0.1t^2 - 2t + 0.2, we want to find when v(t) > 0.
To solve this inequality, we again factor the quadratic equation:
0.1t^2 - 2t + 0.2 > 0
(0.1t - 2)(t + 0.1) > 0
Considering the sign of each factor:
(0.1t - 2) > 0 and/or (t + 0.1) > 0
Solving each inequality and taking their union:
0.1t > 2 and/or t > -0.1
t > 20 or t > -0.1
So, the interval(s) over which the particle is moving forward is t > 20.
c) Similarly, to find the interval(s) over which the particle is moving backwards, we consider when the velocity is negative.
Since v(t) = 0.1t^2 - 2t + 0.2, we want to find when v(t) < 0.
Again, we factor the quadratic equation:
0.1t^2 - 2t + 0.2 < 0
(0.1t - 2)(t + 0.1) < 0
Considering the sign of each factor:
(0.1t - 2) < 0 and/or (t + 0.1) < 0
Solving each inequality and taking their union:
0.1t < 2 and/or t < -0.1
t < 20 or t < -0.1
Since time cannot be negative, we disregard t < -0.1.
Therefore, the interval(s) over which the particle is moving backwards is t < 20.
To find the time(s) at which the particle is not moving, we need to determine the value(s) of t where the particle's velocity is 0.
a) Setting v(t) = 0, we have:
0.1t^2 - 2t + 0.2 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 0.1, b = -2, and c = 0.2. Plugging these values into the formula, we get:
t = (-(-2) ± √((-2)^2 - 4(0.1)(0.2))) / (2(0.1))
Simplifying further:
t = (2 ± √(4 - 0.08)) / 0.2
t = (2 ± √3.92) / 0.2
To find the values of t, we have two cases:
Case 1: t = (2 + √3.92) / 0.2
Case 2: t = (2 - √3.92) / 0.2
Evaluating these cases:
Case 1: t = 10.08 / 0.2 = 50.4 seconds
Case 2: t = -7.92 / 0.2 = -39.6 seconds
Since time cannot be negative in this context, we discard t = -39.6 seconds as an extraneous solution. Therefore, the particle is not moving at t = 50.4 seconds.
b) To determine the interval(s) over which the particle is moving forward, we need to find the values of t where the velocity is positive (v(t) > 0).
For v(t) = 0.1t^2 - 2t + 0.2 > 0, we can set up the quadratic inequality:
0.1t^2 - 2t + 0.2 > 0
To solve this inequality, we can use various methods, such as factoring or the quadratic formula. In this case, let's use the quadratic formula:
Using the same coefficients as before (a = 0.1, b = -2, and c = 0.2), we have:
t = (-(-2) ± √((-2)^2 - 4(0.1)(0.2))) / (2(0.1))
Simplifying further:
t = (2 ± √(4 - 0.08)) / 0.2
t = (2 ± √3.92) / 0.2
We have two cases:
Case 1: t = (2 + √3.92) / 0.2
Case 2: t = (2 - √3.92) / 0.2
Evaluating these cases:
Case 1: t = 10.08 / 0.2 = 50.4 seconds
Case 2: t = -7.92 / 0.2 = -39.6 seconds
Since we are only interested in values of t ≥ 0, we discard t = -39.6 seconds. Therefore, the particle is moving forward for t ≥ 50.4 seconds.
c) Similarly, to determine the interval(s) over which the particle is moving backward, we need to find the values of t where the velocity is negative (v(t) < 0).
For v(t) = 0.1t^2 - 2t + 0.2 < 0, we can set up the quadratic inequality:
0.1t^2 - 2t + 0.2 < 0
Using the same quadratic formula, we have:
t = (-(-2) ± √((-2)^2 - 4(0.1)(0.2))) / (2(0.1))
Simplifying further:
t = (2 ± √(4 - 0.08)) / 0.2
t = (2 ± √3.92) / 0.2
Again, we have two cases:
Case 1: t = (2 + √3.92) / 0.2
Case 2: t = (2 - √3.92) / 0.2
Evaluating these cases:
Case 1: t = 10.08 / 0.2 = 50.4 seconds
Case 2: t = -7.92 / 0.2 = -39.6 seconds
Since we are only interested in values of t ≥ 0, we discard t = -39.6 seconds. Therefore, the particle is not moving backward for any interval.
In summary:
a) The particle is not moving at t = 50.4 seconds.
b) The particle is moving forward for t ≥ 50.4 seconds.
c) The particle is not moving backward for any interval.