Physics, asked by sreejithagutti7646, 1 year ago

An aeroplane travelled a distance of 400 km at an average speed of x km/hr . on return journey, the speed was increased by 40 km/hr. if return journey took 30 in less than the onward journey . find the speed

Answers

Answered by nobel
10
Motion,

We have,
An aeroplane travelled 400 km at an avg. speed of x km/hr. And the return speed was 40 km/hr more than the onward speed.

So the onward speed = x km/hr.
And the return speed = (x + 40) km/hr.

Let the time taken in the onward journey = y hr
And the return journey = z hr.
Also return journey takes 30 min.(30/60=1/2 hr.) less time.

So y - z = 1/2........(1)

Now time = distance / speed
So, y = 400/x hr.........(2)
And, z = 400/(x + 40) hr. .........(3)

Placing the values from the equations 2 and 3 in equation 1 we get,

(400/x) - {400/(x + 40)} = 1/2
or, (400x + 16000 - 400x)/(x+40)x = 1/2
or, 16000×2 = x²-40x
or, x²-40x-32000 = 0
or, x²+200x-160x-32000 = 0
or, x(x+200)-160(x+200) = 0
or, (x+200)(x-160) = 0

So either (x+200) = 0 or,(x-160) = 0
Then x = -200 then x = 160

But speed can't be negative because time is a scalear quantity.

So the onward speed was 160 km/hr and the return speed was (160+40)km/hr = 200km/hr.

That's it
Hope it helped •̀.̫•́✧
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