A boy moves 100m towards North then 70 my towards East then what is its displacement
Is there a formula for displacement?
Displacement is position vector
in this case it is
70 i + 100 j in x y coordinates where i is unit x and j is unit y vectors
its magnitude is sqrt (70^2 + 100^2)
it direction north of east is
tan theta = 100/70
the compass angle east of north is 90 -theta
The body displaces 100m north and then 70m east.
Let Ā (position vector) = 100m due north and Ē(position vector) = 70m due east. Then the displacement is the resultant of the two vectors Ā and Ē.
The angle theta between the two vectors is 90° (refer the position of four cardinal directions i.e.,north , south, east and west.)
Using triangle law of vector addition,
Resultant = Square root (|Ā|² + |Ē|² + 2|Ā||Ē|cos theta )
= Square root ( 100² + 70² + 2(100)(70)cos90°)
● cos 90° = 0
= Square root ( 100² + 70² + 0)
Therefore,
resultant = square root (14900)
= 122.06 or 122.1 m
To find the displacement of the boy, we can use the concept of vector addition. The boy initially moves 100 meters towards the north, which can be represented as a vector pointing upwards. Then, the boy moves 70 meters towards the east, which can be represented as a vector pointing to the right.
To calculate the displacement, we need to add these two vectors together. We can use the Pythagorean theorem to find the magnitude of the resultant vector, and then use trigonometry to find the direction.
Step 1: Calculate the magnitude of the resultant vector:
We have a right-angled triangle with the vertical side representing 100 meters (north) and the horizontal side representing 70 meters (east). Using the Pythagorean theorem (a^2 + b^2 = c^2), we can calculate the length of the hypotenuse (c), which represents the magnitude of the resultant vector.
c = sqrt((100^2) + (70^2))
= sqrt(10000 + 4900)
= sqrt(14900)
≈ 122.02 meters
Step 2: Calculate the direction of the resultant vector:
To find the direction, we can use the inverse tangent function (tan⁻¹) on the ratio of the vertical side (north) to the horizontal side (east) of the triangle.
θ = tan⁻¹(vertical side/horizontal side)
= tan⁻¹(100/70)
≈ 55.01 degrees
So, the displacement of the boy is approximately 122.02 meters in a direction of 55.01 degrees (measured from the north, clockwise).