Question

A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Answer 1

Let the usual speed of the plane be x km/hr.

Usual time = 1500/x
when speed is increased by 100km/hr, then the speed = 1500/x+100
It got late by 1/2 hr.
A.T.Q
1500/x - 1/2 = 1500/x+100
1500/x - 1500/x+100 = 1/2
1500(1/x - 1/x+100) = 1/2
1500( x+100-x/x^2+100x)= 1/2
1500(100/x^2+100x) = 1/2
150000*2 = x^2+100x
300000=x^2+100x
x^2+100x-300000
Using quadratic formula,
a=x^2 b=100 c=-300000
x=-b+-√b^2-4ac/2a
x=-100+-√10000+1200000/2
x=-100+-√1210000/2
x=-100+-1100/2
x=-100+1100/2 or x=-100-1100/2
x=1000/2 or x=-1200/2
x=500km/hr or x=-600km/hr X(neglected)
As the speed cannot be -ve
Therefore, the speed of the plane is 500km/hr.

Answer 2

Given:

Late = 30 min

Distance = 1500 km

Speed increase = 100 km / hr

To find:

The usual speed.

Solution:

By formula,

Speed = Distance / Time  

Since the time is unknown,  

Consider x for time,

Speed original = 1500 / x  

Time taken = ( x - 1 / 2 )  

Therefore,

Substituting the value of x,

Speed new = 1500 / (x - 1/2)  

Speed New - Speed original = 100

( 1500 / ( x - 1 / 2 ) ) - ( 1500 / x ) = 100

150 ( ( 10 / x - 1 / 2 ) - 10 / x = 100  

30 x -30 x +15 = 2x^2 - x

15 = 2x^2 - x

2x^2 - x -15 = 0

Solving,

x = -5/2

x = 3

As time cannot be negative,

The usual time taken by the plane is 3 hr

Speed original = ( 1500 / 3 ) = 500 km / hr

Hence, the usual speed of the plane is 500 km / hr

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