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!
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
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
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
Sure! Let's break down the problem step by step:
a) To find the time(s) at which the particle is not moving, we need to find when the velocity (v(t)) is equal to zero. In other words, we need to solve the equation 0.1t^2 - 2t + 0.2 = 0 for t.
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 0.1, b = -2, and c = 0.2. Plugging those values into the formula, we get:
t = (-(-2) ± √((-2)^2 - 4(0.1)(0.2))) / (2(0.1))
Simplifying the equation further:
t = (2 ± √(4 - 0.08)) / 0.2
t = (2 ± √3.92) / 0.2
t ≈ (2 ± 1.98) / 0.2
So, the two possible values of t when the particle is not moving are:
t ≈ (2 + 1.98) / 0.2 ≈ 19.9 seconds
and
t ≈ (2 - 1.98) / 0.2 ≈ 0.1 seconds
b) To find the interval(s) over which the particle is moving forward, we need to determine when the velocity (v(t)) is positive. This means we want to find the values of t for which v(t) > 0.
We can solve this inequality by setting v(t) > 0:
0.1t^2 - 2t + 0.2 > 0
To solve this quadratic inequality, we can use several methods. One approach is to find the roots of the quadratic equation (where v(t) = 0) and then determine the sign of the expression in between those roots.
Using the quadratic formula, we know the roots of 0.1t^2 - 2t + 0.2 = 0 are approximately t ≈ 0.1 seconds and t ≈ 19.9 seconds, as we found earlier.
To determine the sign of v(t) in the intervals between the roots, we can use test points. We can choose a value of t that is less than 0.1, between 0.1 and 19.9, and greater than 19.9 to test which intervals satisfy v(t) > 0.
For example, if we choose t = 1, we can plug that value into the equation: v(1) = 0.1(1)^2 - 2(1) + 0.2 = -1.7
Since v(1) is negative, the interval before t = 0.1 does not satisfy v(t) > 0.
By choosing test points in the other intervals, we find that the particle is moving forward for t > 19.9 (to infinity) and t < 0.1 (to negative infinity).
So, the interval(s) over which the particle is moving forward can be expressed as:
(-∞, 0.1) U (19.9, ∞)
c) The interval(s) over which the particle is moving backward can be found by determining when the velocity (v(t)) is negative. We want to find the values of t for which v(t) < 0.
Using the same reasoning as in part b), we can conclude that the particle is moving backward for the interval:
(0.1, 19.9)
I hope this explanation helps! Let me know if there's anything else you need assistance with.