A particle is moving on x-y plane. It starts from origin and moves 2m along x-axis. Then it moves another 6√2 m along a direction making an angle of 45° with +ve x-axis in anticlockwise direction. The total displacement of the particle is:
Answers
Answer:
Initial velocity 2
i
^
m/s
Time =2sec
Hence the new velocity will be along x-axis and y-axis (two components)
So according will also have two components
According to along y-axis
Initial velocity =0
i
^
m/s
Final velocity=4sin60°m/s
=4
2
3
=
2
4
3
=
2
4
3
÷2
=
3
m/s
The displacement of the particle is 10 m.
Given: A particle starts from origin and moves 2 m along the x-axis. Then it moves another 6√2 m along a direction making an angle of 45° with +ve x-axis in an anticlockwise direction.
To Find: The total displacement of the particle.
Solution:
- The displacement can be described by the shortest distance between two points.
- Displacement is a vector quantity. It can be positive, negative, or zero.
- Since displacement is a vector quantity, so the final displacement can be calculated by finding the resultant vector. The resultant vector can be found by the formula,
Resultant = √( dx² + dy² ) .......(1)
Where dx = displacement along x-direction, dy = displacement along y-direction.
Coming to the numerical, we are given;
Distance along x-direction = 2 m
It is said that the particle moves another 6√2 m along a direction making an angle of 45° with +ve x-axis,
So, the displacement along x-direction = 6√2 cos 45° = ( 6√2 × ( 1/√2))
= 6 m
And, the displacement along y-direction = 6√2 sin 45° = ( 6√2 × ( 1/√2))
= 6 m
So, the total displacement along x-direction = ( 2 + 6 ) m
= 8 m
The displacement along y-direction = 6 m
So, from (1), we can find the resultant displacement by putting respective values in the formula,
Resultant = √( 6² + 8² )
= √( 36 + 64 )
= √100
= 10 m
Hence, the displacement of the particle is 10 m.
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