Question

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

Answer 1

Answer: U = 600 m/s

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

500km/hr

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