A walks 10 metre in front and 10 metre to the right. Then every time turning to his left, he walks 5, 15 and 15 metre respectively. What is the distance from his initial point ?

examrobotsa's picture
Q: 22 (CAPF/2008)
A walks 10 metre in front and 10 metre to the right. Then every time turning to his left, he walks 5, 15 and 15 metre respectively. What is the distance from his initial point ?

question_subject: 

Maths

question_exam: 

CAPF

stats: 

0,30,9,30,5,4,0

keywords: 

{'distance': [0, 3, 3, 3], 'metre': [0, 3, 4, 3], 'walks': [0, 0, 1, 0], 'initial point': [0, 0, 1, 0], 'front': [1, 0, 4, 7]}

In this question, A starts by walking 10 meters in front and 10 meters to the right. This means that A has moved 10 meters in the north direction and 10 meters in the east direction, forming a right angle.

Then, A turns to his left and walks 5 meters. The direction in which A moves after this turn is north, since he turned left from the east.

Next, A turns to his left again and walks 15 meters. Now, A is moving in the west direction.

Finally, A turns to his left for the last time and walks 15 meters. Now, A is moving south.

To find the distance from A`s initial point, we need to calculate the net displacement made in the north and east directions.

In this case, A has moved 10 meters in the north direction (from the initial point) and 10 meters in the east direction.

Since the question asks for the distance from the initial point, we can consider these two displacements as two sides of a right-angled triangle.

Using the Pythagorean theorem, we can find the distance from the initial point:

Distance = √(10² + 10²)

= √(100 + 100

Practice this on app