A particle undergoes a displacement Δr⃗ of magnitude 54 m in a direction 42 ∘ below the x axis. Express Δr⃗ in terms of the unit

Question

Rashid [163]

2 years ago

10

A particle undergoes a displacement Δr⃗ of magnitude 54 m in a direction 42 ∘ below the x axis. Express Δr⃗ in terms of the unit

vectors x^ and y^.

Physics

2 answers:

Paraphin [41] · Answer 1 · 2023-12-21T04:02:01+03:00

There are several ways to write this vector:
Form 1:
Angle measured clockwise from the x axis
Δr = 54 * (cos (42) x ^ - sin (42) y ^)
Form 2:
Angle measured counterclockwise from the x axis
Δr = 54 * (cos (318) x ^ + sin (318) y ^)
answer
Δr = 54 * (cos (42) x ^ - sin (42) y ^)
Δr = 54 * (cos (318) x ^ + sin (318) y ^)

Rom4ik [11] · Answer 2 · 2023-12-21T05:45:47+03:00

The displacement of the particle in terms of its unit vectors is given by $\boxed{\Delta r=(40.13\text{ m})\,\hat x-(36.13\text{ m})\hat y}$ .

Explanation:

Given:

The magnitude of displacement of the particle is $54\text{ m}$ .

The angle made by the particle with the X-axis is $42^\circ$ in negative direction.

Concept:

The displacement of a body is defined as the shortest distance between the initial and the final position of the body. The displacement of a body includes the direction along with the magnitude of displacement of the body. Therefore, displacement is termed as the vector quantity.

The displacement of the particle is in the positive X-direction but making an angle of $42^\circ$ below the X-axis. Therefore, the overall displacement of the particle is in the positive direction of X-axis and negative direction of Y-axis.

The displacement of the particle is represented as:

$\boxed{\Delta r=A\,cos\,42^\circ\,\hat x\,-A\,sin\,42^\circ\,\hat y}$