A freight train made a trip to a repair. On the trip there it travelers 25 mph and on the return trip it went 20 mph. How long d
1 answer:
Answer:
Time taken by train in onward journey = 12 hours.
Step-by-step explanation:
Given:
Speed of train making a trip to a repair = 25 mph
Speed of train on return trip = 20 mph
Time taken for return trip = 15 hours
To find the time taken on the on wards trip.
Solution:
The distance traveled by the train on the trip and return trip is the same as the y are of same trips in opposite directions.
Distance can be calculated by using the data for the return trip.
Distance=
Distance=
Speed of train for on ward trip = 25 mph
Time taken =
Time taken =
Thus, time taken by train in onward journey = 12 hours.
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Answer:
The highest altitude that the object reaches is 576 feet.
Step-by-step explanation:
The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be , the first and second derivatives are, respectively:
First Derivative
Second Derivative
Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:
(Critical value)
The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:
The highest altitude that the object reaches is 576 feet.