A flea moves along the line y = -3 according to the equation: x = t^3 - 9t^2 + 24t where x is its x coordinate at time t secs.
Determine:
1) the position, velocity, and acceleration of the flea at time 3 secs.
Answer: position - 3^3 - 9(3)^2 + 24(3) = 18 units/second
velocity - 3(3)^2 - 18(3) + 24 = -3 units/second
acceleration: 6(3) - 18 = 0 units/second^2
2) the time(s) when the flea is at rest
Answer: 3t^2 - 18t + 24 = 0
3(t^2 - 6t + 8) = 0
3(t-4)(t-2) = 0
t = 4 secs or t = 2 secs
3) the interval(s) when the flea is moving right and the interval(s) when he is moving left
4) the leftmost and rightmost positions of the flea
5) the net displacement of the flea from time 0 sec. to time 5 sec.
6) the total distance traveled by the flea during the 1st 5 seconds
Answer: 12 units
Are my answers correct? I'm not sure how to find the answers to the other ones.
#1 and #2 look ok (except that x(3) = 18 units, not units/s)
#3 flea moving right when x'(t) > 0 (s increasing)
#4 evaluate x(t) at the extrema from part 2.
Check with this graph:
http://www.wolframalpha.com/input/?i=t^3+-+9t^2+%2B+24t
#5 x(5)-x(0)
#6 ∫[0,5] (1+(x')^2) dt
= ∫[0,5] √(1+(3t^2-18t+24)^2) dt
= 29 units
Not sure where the 12 came from.
Let's go through each question one by one:
1) To determine the position, velocity, and acceleration of the flea at time 3 seconds, we need to substitute t = 3 seconds into the given equation for x.
The position at time t is given by x = t^3 - 9t^2 + 24t. So, plugging in t = 3, we get x = (3^3) - 9(3^2) + 24(3) = 18 units.
To find the velocity, we need to take the derivative of the position function with respect to time. The derivative of x(t) = t^3 - 9t^2 + 24t is given by dx/dt = 3t^2 - 18t + 24. Substituting t = 3, we get dx/dt = 3(3^2) - 18(3) + 24 = -3 units/second.
To find the acceleration, we need to take the derivative of the velocity function with respect to time. The derivative of dx/dt = 3t^2 - 18t + 24 is given by d^2x/dt^2 = 6t - 18. Substituting t = 3, we get d^2x/dt^2 = 6(3) - 18 = 0 units/second^2.
So, the position of the flea at time 3 seconds is 18 units, the velocity is -3 units/second, and the acceleration is 0 units/second^2.
2) To find the time(s) when the flea is at rest, we need to find the values of t for which the velocity is 0. We already found the velocity function to be dx/dt = 3t^2 - 18t + 24. Setting this equal to 0, we get:
3t^2 - 18t + 24 = 0.
This is a quadratic equation, which we can solve using factoring, completing the square, or the quadratic formula. Factoring gives us 3(t-4)(t-2) = 0. So, the flea is at rest at t = 4 seconds or t = 2 seconds.
3) To determine the intervals when the flea is moving right and when it is moving left, we can look at the sign of the velocity. We already found the velocity function to be dx/dt = 3t^2 - 18t + 24.
To find when the flea is moving right, we need to find the intervals where the velocity is positive (greater than 0). Setting the velocity function greater than 0, we get 3t^2 - 18t + 24 > 0. Solving this inequality will give us the intervals when the flea is moving right.
Similarly, to find when the flea is moving left, we need to find the intervals where the velocity is negative (less than 0). Setting the velocity function less than 0, we get 3t^2 - 18t + 24 < 0.
To solve these inequalities, we need to factor or use the quadratic formula.
4) The leftmost position of the flea can be found by determining the minimum value of the position function x. To find this, we can use calculus by finding the critical points (where the derivative is 0) and checking which one gives the smallest x value. Alternatively, we can use the minimum value of the quadratic function.
The rightmost position of the flea can be found by determining the maximum value of the position function x. Again, we can use calculus or the maximum value of the quadratic function.
5) The net displacement of the flea from time 0 seconds to time 5 seconds can be found by calculating the difference in the position function x(5) - x(0), where x(t) is the position function. Substitute t = 5 seconds and t= 0 seconds into the position function and calculate the difference.
6) The total distance traveled by the flea during the first 5 seconds can be found by adding up the absolute values of the distances traveled during each interval when the flea is moving right or left. Calculate the distances traveled during each interval separately and sum them up.
To check whether your answers are correct, you can go through each step carefully and double-check your calculations. If you are still unsure about any particular question, let me know and I can guide you further.