Question

A pilot flies distance of 800 km in some average speed. He could have saved 40 minutes by increasing the average speed by 40km/h . Find the actual speed of the aircraft.

Respected Experts,

The answer o fthe above question has been posted earlier by Neha Madam, but I could not understand it. I request the solution of the above problem in a more simplified manner. Thanks for your great!!!!!! support.

Answer 1

The relation between speed v, distance d, and time t is defined by the formula:

$v = \frac{d}{t}$

We can rearrange the formula to instead give the time t that is needed to cover a distance d travelling at speed v:

$t = \frac{d}{v}$

The problem states that by increasing v by 40 km/h, the same 800 km distance can be covered in 40 minutes less time. We have to convert 40 mins to hours to unify the unit of our calculation. In this case, 40 mins is 2/3 of an hour. Expressing the problem using the formulas we know so far:

$t _{diff} = \frac{d}{v} - \frac{d}{v+40}$

where $t_{diff}$  is the time difference between the two speed.

Plug in the known values:

$\frac{2}{3} = \frac{800}{v} - \frac{800}{v + 40}$

Multiply both sides by $v+40$:

$\frac{2}{3} (v+40) = (\frac{800}{v} - \frac{800}{v + 40})v+40$

After a bit of simplification:

$\frac{2}{3}v + \frac{80}{3} = \frac{32000}{v}$

Multiply both sides by v you get:

$\frac{2}{3} v^{2} + \frac{80}{3} v = 32000$

Multiplying both side by 3 to get rid of the fractions you get:

$2v^{2} + 80v = 96000$

Simplifying:

$v^{2} + 40v = 48000$

To solve for v, we need to solve the quadratic equation.

There are many ways of doing this or you can use an online quadratic calculator.

There are two possible solutions to this quadratic equation:

$v = 200,\\v = -240$

Since we're talking of speed here, the value can't be negative.

Therefore, the actual speed of the aircraft is 200 km/h.

Please note that the formal way of doing any calculations (especially in physics) involving units of measures (km, km/h, h, etc.) is to ALSO write those units in every step of the calculation.