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A ship A is moving westwards with a speed of 10kmh−1 and a ship B 100 km South of A, is moving northwards with a speed of 10kmh−1. The time after which the distance between them becomes shortest, is
- a)5√2h
- b)10√2h
- c)0 h
- d)5 h
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A ship A is moving westwards with a speed of 10kmh−1 and a ship B 100...
To find the time after which the distance between the two ships becomes shortest, we can use the concept of relative velocity.
Relative velocity is the vector difference between the velocities of two objects. In this case, we have two ships, A and B, moving in different directions with different velocities.
Let's consider the initial positions of the two ships as shown below:
Ship A:
Initial position: (0, 0) (Assuming it starts from the origin)
Velocity: 10 km/h westwards (negative x-direction)
Ship B:
Initial position: (0, -100) (100 km south of A)
Velocity: 10 km/h northwards (positive y-direction)
We need to find the time when the distance between the two ships is shortest. To do that, we can consider a point P at any arbitrary time t, where the distance between A and B is minimum.
Let's assume that at time t, the position of ship A is (x, 0) and the position of ship B is (0, -100 + 10t). The x-coordinate of ship A remains the same as it moves only in the x-direction, while the y-coordinate of ship B changes with time as it moves in the y-direction.
To find the distance between A and B at time t, we can use the distance formula:
Distance = √((x - 0)^2 + (0 - (-100 + 10t))^2)
= √(x^2 + (100 - 10t)^2)
Now, we need to minimize this distance. To find the minimum, we can differentiate the distance function with respect to time t and equate it to zero:
d(Distance)/dt = 0
Differentiating the distance function with respect to t, we get:
-20(100 - 10t) = 0
-2000 + 200t = 0
200t = 2000
t = 10 hours
Therefore, the time after which the distance between the two ships becomes shortest is 10 hours.
Relative velocity is the vector difference between the velocities of two objects. In this case, we have two ships, A and B, moving in different directions with different velocities.
Let's consider the initial positions of the two ships as shown below:
Ship A:
Initial position: (0, 0) (Assuming it starts from the origin)
Velocity: 10 km/h westwards (negative x-direction)
Ship B:
Initial position: (0, -100) (100 km south of A)
Velocity: 10 km/h northwards (positive y-direction)
We need to find the time when the distance between the two ships is shortest. To do that, we can consider a point P at any arbitrary time t, where the distance between A and B is minimum.
Let's assume that at time t, the position of ship A is (x, 0) and the position of ship B is (0, -100 + 10t). The x-coordinate of ship A remains the same as it moves only in the x-direction, while the y-coordinate of ship B changes with time as it moves in the y-direction.
To find the distance between A and B at time t, we can use the distance formula:
Distance = √((x - 0)^2 + (0 - (-100 + 10t))^2)
= √(x^2 + (100 - 10t)^2)
Now, we need to minimize this distance. To find the minimum, we can differentiate the distance function with respect to time t and equate it to zero:
d(Distance)/dt = 0
Differentiating the distance function with respect to t, we get:
-20(100 - 10t) = 0
-2000 + 200t = 0
200t = 2000
t = 10 hours
Therefore, the time after which the distance between the two ships becomes shortest is 10 hours.
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Community Answer
A ship A is moving westwards with a speed of 10kmh−1 and a ship B 100...
Given situation is shown in the figure
OS = 100 km
OP = Shortest distance
Relative velocity between A and B is
The time after which distance between them equals to OP is given by
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A ship A is moving westwards with a speed of 10kmh−1 and a ship B 100 km South of A, is moving northwards with a speed of 10kmh−1. The time after which the distance between them becomes shortest, isa)5√2hb)10√2hc)0 hd)5 hCorrect answer is option 'D'. Can you explain this answer?
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