An aeroplane covers a distance of 1500km in a certain time at a certain speed.After increasing the speed by 100km/hr, it covers the same distance in a time which is half an hour less than the previous time. Find the previous speed of the aeroplane.
this is from quadratic equations CBSE grade 10
please answer ASAP

Answers

Answer 1

Answer: U = 600 m/s

Step-by-step explanation:

Given that an aeroplane covers a distance of 1500km in a certain time t at a certain speed U.

After increasing the speed by 100km/hr, that is, V = U + 100 it covers the same distance in a time which is half an hour less than the previous time. That is t2 = t - 0.5.

From the first statement

Speed = distance/ time

Distance = speed × time

1500 = Ut

Make t the subject of the formula

t = 1500/U ..... (1)

From the second statement

Distance = speed × time

1500 = (U + 100) × ( t - 0.5 )

Open the bracket

1500 = Ut - 0.5U + 100t - 50

Collect the like terms

1550 = Ut - 0.5U + 100t .... (2)

Substitutes equation 1 into 2

1550 = 1500U/U - 0.5U + 100(1500/U)

1550 = 1500 - 0.5U + 150000/U

1550 - 1500 = (150000 - 0.5U^2)/U

Cross multiply

50U = 150000 - 0.5U^2

0.5U^2 + 50U - 150000 = 0

Divide all by 0.5

U^2 + 100U - 300000 = 0

Using completing the square method

U^2 + 100U = 300000

U^2 + 100U + 50^2 = 300000 + 50^2

(U + 50)^2 = 302500

U + 50 = sqrt(302500)

U + 50 = +/-(550)

U = 50 + 550 or 50 - 550

U = 600 or - 500

Since U is of the same direction, it is

positive. Therefore, the previous speed of the aeroplane is 600 m/s

Answer 2

Answer:

500km/hr

Step-by-step explanation:

The formula for Speed (km/hr) = Distance / Time

Where S = Speed

D = Distance

T = Time

S = D/T

From the question, the aeroplane covered a distance of 1500 km

S = 1500/ T

ST = 1500

Time taken( T ) = 1500/S ......... Equation 1

We are told from the question that speed was increases by 100km/hr, it covers the same distance in a time which is half an hour less than the previous time

This is expressed mathematically as:

The new speed =S + 100km/hr

The new Time taken = 1500/ S - 1/2....... Equation 2

Also since Time = Distance / Speed

The new Time taken also = 1500/ S + 100 ......... Equation 3

Step 1

We would simplify Equation 2:

Time = 1500/S - 1/2

Find the Lowest common multiple = 2

Time = (2 × 1500 - S )/ 2S

Time = 3000 - S / 2S ......... Equation 4

Step 2

Equate Equation 4 and 3 together since they are both equal to time taken

1500/ S + 100 = 3000 - S / 2S

We cross multiply

2S × 1500 = (S + 100) ( 3000 - S)

3000S = 3000S - S² + 300000 - 100S

3000S - 3000S + S² - 300000 + 100S = 0

S² + 100S - 300000 = 0

Step 3

We solve for S = Speed by using factorisation method.

S² + 100x - 300000 = 0

S² - 500S + 600S - 300000 = 0

(S² - 500S) + (600S - 300000) = 0

S(S - 500) + 600(S - 500) = 0

(S - 500) (S + 600) = 0

S - 500 = 0, S = 500

S + 600 = 0 , S = -600

Our answer cannot be in negative form, hence, the previous speed of the aeroplane = 500km/hr


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