Answer:
Difference = 1.83 hours
Explanation:
Let the two trips be A and B respectively.
Given the following data;
Distance A = 2000 miles
Speed A = 400 miles
Distance B = 2000 miles
Speed B = 480 miles per hour
To find the time spent;
a. For trip A.
Time A = distance A/speed A
Time A = 2000/400
Time A = 5 hours
b. For trip B.
Time B = distance B/speed B
Time B = 2000/480
Time B = 4.17 hours
Next, we would find the difference between the two trips.
Difference = Time A - Time B
Difference = 5 - 4.17
Difference = 1.83 hours
Therefore, the return trip was much quicker by 1.83 hours.
The return trip of the airplane will be quicker by 1.84 hours
It is given that
Distance covered by the plane D= 2000 miles
The average speed of the plane
[tex]V_1=400\ \frac{Miles }{hour }[/tex]
The Return Average speed of the plane
[tex]V_2= 480\ \frac{Miles}{hour}[/tex]
Now we will first find the time taken by the airplane during the first visit.
[tex]V_1=\dfrac{Distance }{Time }[/tex]
[tex]Time = \dfrac{Distance }{V_1}[/tex]
Now by putting the values in the formula
[tex]T_1=\dfrac{2000}{400} =5\ hours[/tex]
Now for the return trip time taken by the plane
[tex]T_2=\dfrac{Distance }{V_2}[/tex]
[tex]T_2= \dfrac{2000}{480}[/tex]
[tex]T_2= 4.16\ hours[/tex]
Now the difference between the time will be
[tex]=T_1-T_2=5-4.16=0.86=1.84 \ Hours[/tex]
Thus the return trip of the airplane will be quicker by 1.84 hours
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