
A bus appears to go with a speed of \[25{km}/{hr}\;\]to a car driver, driving at the rate \[7{km}/{hr}\;\] northwards. If the bus actually travels in the east direction, its speed is
\[A.\,24{km}/{hr}\;\]
\[B.\,23{km}/{hr}\;\]
\[C.\,26{km}/{hr}\;\]
\[D.\,30{km}/{hr}\;\]
Answer
568.8k+ views
Hint: This question is based on the concept of the directions. The resultant velocity is the solution of the answer along the north-east direction, as the car is moving along the north direction and the bus is considered to be moving along the east direction. The resultant velocity is the root of the sum of the squares of the velocities of the bus (according to the car driver) and car.
Formula used:
\[V_{R}^{2}=V_{x}^{2}+V_{y}^{2}\]
Complete answer:
The resultant velocity is given by the formula as follows.
\[V_{R}^{2}=V_{x}^{2}+V_{y}^{2}\]
Where \[{{V}_{x}}\] represents the vector along the x-axis direction and \[{{V}_{y}}\] represents the vector along the y-axis direction.
From the data, we have the data as follows.
To a car driver, the bus appeared to go with a speed of \[25{km}/{hr}\;\], but the direction is not mentioned. So, \[{{V}_{NE}}=25{km}/{hr}\;\]
The car driver is driving the car at the rate\[7{km}/{hr}\;\] northwards. So, \[{{V}_{N}}=7{km}/{hr}\;\]
The question is to find the speed of the bus, if considered, the direction of the bus is east.
So, we have,
The car is moving in the north direction and the bus is considered to be moving in the east direction, thus, the resultant velocity will be along the north-east direction. Thus, we have the formula as follows.
\[V_{NE}^{2}=V_{E}^{2}+V_{N}^{2}\]
Substitute the given values in the above equation.
\[\begin{align}
& {{25}^{2}}=V_{E}^{2}+{{7}^{2}} \\
& \Rightarrow V_{E}^{2}={{25}^{2}}-{{7}^{2}} \\
\end{align}\]
Continue the further calculation.
\[\begin{align}
& V_{E}^{2}=625-49 \\
& \Rightarrow {{V}_{E}}=24{km}/{hr}\; \\
\end{align}\]
\[\therefore \] The value of the speed of the bus, considering the actual direction of the bus to be along the east direction is \[24{km}/{hr}\;\] .
Thus, the option (A) is correct.
Note:
The question is a bit confusing while considering the directions. As the two of the three directions are given, so it is obvious that, the resultant velocity will have the direction in between the other two velocities in some cases and the resultant will be in between but tilted.
Formula used:
\[V_{R}^{2}=V_{x}^{2}+V_{y}^{2}\]
Complete answer:
The resultant velocity is given by the formula as follows.
\[V_{R}^{2}=V_{x}^{2}+V_{y}^{2}\]
Where \[{{V}_{x}}\] represents the vector along the x-axis direction and \[{{V}_{y}}\] represents the vector along the y-axis direction.
From the data, we have the data as follows.
To a car driver, the bus appeared to go with a speed of \[25{km}/{hr}\;\], but the direction is not mentioned. So, \[{{V}_{NE}}=25{km}/{hr}\;\]
The car driver is driving the car at the rate\[7{km}/{hr}\;\] northwards. So, \[{{V}_{N}}=7{km}/{hr}\;\]
The question is to find the speed of the bus, if considered, the direction of the bus is east.
So, we have,
The car is moving in the north direction and the bus is considered to be moving in the east direction, thus, the resultant velocity will be along the north-east direction. Thus, we have the formula as follows.
\[V_{NE}^{2}=V_{E}^{2}+V_{N}^{2}\]
Substitute the given values in the above equation.
\[\begin{align}
& {{25}^{2}}=V_{E}^{2}+{{7}^{2}} \\
& \Rightarrow V_{E}^{2}={{25}^{2}}-{{7}^{2}} \\
\end{align}\]
Continue the further calculation.
\[\begin{align}
& V_{E}^{2}=625-49 \\
& \Rightarrow {{V}_{E}}=24{km}/{hr}\; \\
\end{align}\]
\[\therefore \] The value of the speed of the bus, considering the actual direction of the bus to be along the east direction is \[24{km}/{hr}\;\] .
Thus, the option (A) is correct.
Note:
The question is a bit confusing while considering the directions. As the two of the three directions are given, so it is obvious that, the resultant velocity will have the direction in between the other two velocities in some cases and the resultant will be in between but tilted.
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