Study the information given below carefully and answer the questions that follow: Kartick starts moving in a car from point A. He moves 30 km to the south and then turns left from point B and moves 50 km till point C. He then turns right and moves 60 km and then turns left from point D and moves 50 km till point E. He then turns right from point E and moves 20 km and then turns right from point F. He moves 150 km and then turns right from point G and moves 80 km till point H and stopped.
What is the shortest distance between point H and point F?
To find the shortest distance between point H and point F, we first need to carefully trace Kartick's path from the starting point A, noting down each turn and the distance covered. This type of problem often involves visualizing the movement and then applying geometric principles, specifically the Pythagorean theorem for the shortest distance between two points in a plane.
Let's break down Kartick's movement step-by-step to understand the complete journey:
To accurately determine the shortest distance between point H and point F, it is beneficial to represent the positions of these points using a coordinate system. Let's assume the starting point A is at the origin \((0,0)\). We will track the X-coordinate (East/West) and Y-coordinate (North/South) for each significant point in Kartick's journey. Positive X will represent East, negative X will represent West. Positive Y will represent North, negative Y will represent South.
| Point | X-coordinate (km) | Y-coordinate (km) | Explanation of Movement |
|---|---|---|---|
| A | \(0\) | \(0\) | Starting position |
| B | \(0\) | \(-30\) | Moves 30 km South from A |
| C | \(0 + 50 = 50\) | \(-30\) | Moves 50 km East (left turn from South) from B |
| D | \(50\) | \(-30 - 60 = -90\) | Moves 60 km South (right turn from East) from C |
| E | \(50 + 50 = 100\) | \(-90\) | Moves 50 km East (left turn from South) from D |
| F | \(100\) | \(-90 - 20 = -110\) | Moves 20 km South (right turn from East) from E |
| G | \(100 - 150 = -50\) | \(-110\) | Moves 150 km West (right turn from South) from F |
| H | \(-50\) | \(-110 + 80 = -30\) | Moves 80 km North (right turn from West) from G |
From our detailed coordinate tracking, we have the precise locations for the points we need to analyze:
The shortest distance between any two points in a two-dimensional plane can be calculated using the distance formula, which is directly derived from the Pythagorean theorem. If we have two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(D\) between them is given by:
\[ D = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \]
Let's apply this formula using the coordinates of point F \((100, -110)\) and point H \((-50, -30)\).
Here, \(x_1 = 100\), \(y_1 = -110\) (for point F) and \(x_2 = -50\), \(y_2 = -30\) (for point H).
Substituting these values into the distance formula: \[ D_{HF} = \sqrt{ (-50 - 100)^2 + (-30 - (-110))^2 } \] \[ D_{HF} = \sqrt{ (-150)^2 + (-30 + 110)^2 } \] \[ D_{HF} = \sqrt{ (-150)^2 + (80)^2 } \]
Now, we calculate the squares of the differences: \[ (-150)^2 = 22500 \] \[ (80)^2 = 6400 \]
Adding these values: \[ D_{HF} = \sqrt{ 22500 + 6400 } \] \[ D_{HF} = \sqrt{ 28900 } \]
Finally, we find the square root of 28900: We know that \(17^2 = 289\), so \(\sqrt{289} = 17\). Thus, \(\sqrt{28900} = \sqrt{289 \times 100} = \sqrt{289} \times \sqrt{100} = 17 \times 10 = 170\).
Therefore, the shortest distance between point H and point F is 170 km.
A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s2 and neglecting air resistance will give :
A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
Consider the following statements:
1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :