Question

A plane left 30 min later from the schedule time and in order to reach the destination 1500 km on the time if had increase the speed 250 km/h from the usual speed, its usual speed is:
A. 700 kmph
B. 750 kmph
C. 250 kmph
D. 350 kmph
E. 500 kmph

Answer 1

Hint:Speed is the rate at which an object moves. It’s a very basic concept in motion and all about how fast or slow an object can move. Speed is simply Distance divided by the time where Distance is directly proportional to Velocity when time is constant. Problems related to Speed, Distance, and Time, will ask you to calculate for one of three variables given.
Use the time distance and speed relation formula:
Distance = Time × speed


Complete step by step solution:
Speed is a measure of how quickly an object moves from one place to another. It is equal to the distance traveled divided by the time. It is possible to find any of these three values using the other two.
Let the usual speed of plane is x km/h then
New speed = (x + 250) km/h
According to the distance formula
\[\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}\]
Plane left 30 min or ½ hours later than the schedule time (given)
According to the questions
\[\dfrac{1}{2}\Rightarrow \dfrac{1500}{x}-\dfrac{1500}{\left( x+250 \right)}\]
By Cross multiplication, we get
\[\dfrac{1}{2}=\dfrac{1500\left( x+25 \right)-1500x}{x\left( x+250 \right)}\]
Again by cross multiplication
\[x\left( x+250 \right)=2\left\{ 1500x+1500\times 250 \right\}\]
\[{{x}^{2}}+250x=750,000\]
\[{{x}^{2}}+250x-750,000=0\]
Factorise the 250 x such that multiplication of factor is 750,000 and add/subtraction is 25 x.
\[{{x}^{2}}+1000x-750x-750000=0\]
\[x\left( x+1000 \right)-750\left( x+1000 \right)=0\]
\[\left( x-750 \right)\left( x+1000 \right)=0\]
\[x=750\]
\[x=1000\]
\[x\ne -1000\]speed can’t be negative
Hence \[x=750km/h\]
Thus usual speed of the plane is \[750km/h.\]
Therefore,option B is the right answer

Note:Make sure to convert the values given into suitable units before substituting in the formula. Also write the appropriate unit associated in the final answer.