A plane flew a distance of 4800 km. On the return trip the speed is decreased 200 by km/h and the flight takes 2 hours longer. What's the speed of the return flight. 

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

Let the speed of the plane on the return journey be S.

As the distance traveled is 4800, the time taken `t = 4800/S` .

On the forward journey, the plane flies at a faster speed of S + 200 km/h and the time is 2 hours lesser.

=> `t - 2 = 4800/(S + 200)`

Substitute `t = 4800/s`

=> `4800/S - 2 = 4800/(S + 200)`

=> `(4800 - 2S)(S + 200) = 4800S`

=> `4800S - 2S^2 + 960000 - 400S = 4800S`

=> `2S^2 + 400S - 960000 = 0`

=> `S^2 + 200S - 480000 = 0`

=> `S^2 + 800S - 600S - 480000 = 0`

=> `S(S + 800) - 600(S + 800) = 0`

=> `(S - 600)(S + 800) = 0`

=> S = 600 and S = -800

As the speed cannot be negative, only S = 600 is considered.

The speed of the plane on the return flight is 600 km/h

quantatanu's profile pic

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted on

Let us consider:

Initial speed = V_i


Initial time of flight = T_i


Returning speed = V_r

Returning time of flight = T_r


Given that speed decreased so

V_i > V_r, hence:

V_i - V_r = 200 -------------(1)

And as speed decreased during return so return time of flight

must increase so T_r > T_i, jence:

T_r -T_i = 2  ---------------(2)

V_i T_i = V_r T_r = 4800 -----------------(3)

Therefore:

T_i = 4800/V_i

T_r = 4800/V_r

                                          By (2)

T_r -T_i = 2

=> 4800/V_r - 4800/V_i  = 2

=> (1/V_r - 1/V_i) = 1/2400

=>(V_i - V_r) = V_i * V_r /2400

=> 200 = V_i * V_r /2400     [by (1)]

=> V_i * V_r = 480000

=> V_i = 480000/V_r

 

Putting V_i in (1), we get


480000/V_r - V_r = 200

=> 480000 - (V_r)^2 = 200 * V_r

=> (V_r)^2 +200* V_r - 480000 = 0

V_r = either [-200 + Sqrt( 40000 + 1920000)]/2

      = [-200 + 1400]/2

      = 600

      or

        [-200 - 1400]/2

      = - 800 but this is unacceptible as I was solely working with

                  the speeds (that is magnitude) an cannot be -ve !

So Returning velocity = 600 km/h

   

najm1947's profile pic

najm1947 | Elementary School Teacher | (Level 1) Valedictorian

Posted on

Let the speed of return flight be v km/h

The time for return flight = 4800/v hours

The speed of first leg of flight = v+200 km/h

The time for first leg of flight = 4800/(v+200)

The difference in time = 2 hours

4800/v = 4800/(v+200) + 2, multiplying both sides by v(v+200)

4800(v+200) = 4800v +2v(v+200)

960000 = 2v^2 + 400v

2v^2 + 400v - 960000 = 0

The equation can be solved for 2 values of v as:

v=[-b+sqrt(b^2-4ac)]/2a and v=[-b+sqrt(b^2-4ac)]/2a

v = [-400+sqrt(400^2-4*2*(-960000))]/2*2 = 600km/h and

v = [-400-sqrt(400^2-4*2*(-960000))]/2*2 = -800km/h

as the second value is negative hence we have to take the return flight speed as 600 km/h

The speed of return flight is 600km/h

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