Answer
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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.
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.
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