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A plane loaded with cargo travels 2000 miles at an average speed of 400 Miles per hour. On the return trip on the same route it travels at an average speed of 480 miles per hour. How much quicker was the return trip

Sagot :

Lanuel

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

What will be the time taken by the airplane on the second trip?

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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