Answer:
direction is given as
South of West
Explanation:
Part b)
displacement is given as
now we will have
total displacement is given as
direction is given as
South of West
The position of the particle is given by:
x(t) = t³ - 12t² + 21t - 9
Differentiate x(t) with respect to t to find the velocity x'(t):
x'(t) = 3t² - 24t + 21
Differentiate x'(t) with respect to t to find the acceleration x''(t):
x''(t) = 6t - 24
<span>b. becomes weaker
try this </span>
Answer:at 21.6 min they were separated by 12 km
Explanation:
We can consider the next diagram
B2------15km/h------->Dock
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B1 at 20km/h
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V
So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.
Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.
Answer:
Atomic name is your answer.
