x = 2t^3 - 21t^2 + 60t - 30 is the motion of an ant where x is his location on the number line at time t seconds.
1) When is he going to the left?
Answer: between 3 to 5 seconds
2) When is he going to the right?
Answer: between 0 to 2 seconds
3) What is his net displacement after 4 seconds?
Answer: 28 units
4) How far has he actually traveled during that time?
Answer: 50 units
Are my answers correct?
dx/dt = 6 t^2 - 42 t + 60
= 6 (t^2 - 7 t + 10)
= 6 (t-2)(t-5)
negative from 2 to 5 (Not 3 to 5)
so then would it be positive from 0 to 1?
I assume that we do not do negative time so
v is positive for 0<t<2 and for t >5
when t = 0, x = -30
when t = 2, x = 22
when t = 4, x = 2
so 28 from 0 to 4 (part 3)agree
part 4 +52-20 = 32
part 4 use absolute value of distance
is 52 + 20 = 72
ans part 3 is from - 30 to + 2 = +32
positive v from 0 to 2
negative v from 2 to 5
positive v from 5 to infinity
parabola holds water, crosses below x axis between 2 and 5
vertex at t = 3.5
To determine whether your answers are correct, we need to analyze the given equation: x = 2t^3 - 21t^2 + 60t - 30. This equation represents the motion of an ant on the number line at different time points.
1) To determine when the ant is going to the left, we look for the intervals where the velocity (x') is negative. Calculate the derivative of x with respect to t to find the velocity function: x' = 6t^2 - 42t + 60.
To find the leftward motion, we need to solve the inequality x' < 0. Factoring the quadratic equation, we get (3t - 5)(2t - 6) < 0. The critical points are t = 5/3 and t = 3.
The intervals where the ant is going to the left are (5/3, 3), which means the ant is going to the left between 3 and 5 seconds. Hence, your answer is correct.
2) To determine when the ant is going to the right, we need to find the intervals when the velocity (x') is positive. We already have the velocity function x' = 6t^2 - 42t + 60 from the previous derivation. Now we solve the inequality x' > 0. Factoring the quadratic, we get (3t - 10)(2t - 6) > 0. The critical points are t = 10/3 and t = 3.
The intervals where the ant is going to the right are (-∞, 10/3) and (3, ∞), which means the ant is going to the right between 0 and 2 seconds. Hence, your answer is correct.
3) To find the ant's net displacement after 4 seconds, we substitute t = 4 into the equation x = 2t^3 - 21t^2 + 60t - 30:
x = 2(4)^3 - 21(4)^2 + 60(4) - 30
x = 128 - 336 + 240 - 30
x = 2 units.
Therefore, the ant's net displacement after 4 seconds is 2 units. Your answer, which states 28 units, is incorrect.
4) To determine how far the ant has actually traveled during the given time, we calculate the total distance traveled, disregarding direction. We find the integral of the absolute value of the velocity function over the interval [0, 4]. The velocity function is x' = 6t^2 - 42t + 60, as derived earlier.
∫|x'|dt from 0 to 4 = ∫(6t^2 - 42t + 60)dt from 0 to 4
= [2t^3 - 21t^2 + 60t] from 0 to 4
= (2(4)^3 - 21(4)^2 + 60(4)) - (2(0)^3 - 21(0)^2 + 60(0))
= 100 - 336 + 240 - 0 + 0
= 4 units.
Therefore, the ant has traveled a distance of 4 units during the given time. Your answer, which states 50 units, is incorrect.
To summarize, 1) and 2) are correct, while 3) and 4) are incorrect.