Question

A particle undergoes three consecutive displacements given by vectors →D1=(3.0ˆi−4.0ˆj−2.0ˆk)mm, →D2=(1.0ˆi−7.0ˆj+4.0ˆk)mm, and →D3=(−7.0ˆi+4.0ˆj+1.0ˆk)mm. (a) Find the resultant displacement vector of the particle. (b) What is the magnitude of the resultant displacement? (c) If all displacements were along one line, how far would the particle travel?

a) Resultant displacement vector: (-3i - 7j + 3k) mm, Magnitude: 8.602 mm, Travel distance: 10 mm
b) Resultant displacement vector: (-3i - 7j + 3k) mm, Magnitude: 8.774 mm, Travel distance: 12 mm
c) Resultant displacement vector: (-4i - 7j + 3k) mm, Magnitude: 9.110 mm, Travel distance: 12 mm
d) Resultant displacement vector: (-2i - 6j + 3k) mm, Magnitude: 7.071 mm, Travel distance: 8 mm

Answer 1

The resultant displacement vector of the particle is (-3.0i - 7.0j + 3.0k) mm. The magnitude of this vector is approximately 8.185 mm. If all displacements were along one line, the travel distance would be 33.0 mm.

Step-by-step explanation:

To find the resultant displacement vector of the particle, we need to add the given displacement vectors: →D1, →D2, and →D3. By adding the i, j, and k components of each displacement vector, we get:

→D_resultant = (3.0i - 4.0j - 2.0k) + (1.0i - 7.0j + 4.0k) + (-7.0i + 4.0j + 1.0k)

• D_resultant_x = 3.0 + 1.0 - 7.0 = -3.0i
• D_resultant_y = -4.0 - 7.0 + 4.0 = -7.0j
• D_resultant_z = -2.0 + 4.0 + 1.0 = 3.0k

Therefore, the resultant displacement vector is (-3.0i - 7.0j + 3.0k) mm.

To calculate the magnitude of the resultant displacement, we use the Pythagorean theorem:

Magnitude = √((-3.0)² + (-7.0)² + (3.0)^2) = √(9 + 49 + 9) = √67 ≈ 8.185 mm

For part (c), if all the displacements were along one line, the total travel distance would simply be the sum of the magnitudes of the individual displacement vectors, regardless of direction:

Travel distance = |3.0| + |4.0| + |2.0| + |1.0| + |7.0| + |4.0| + |7.0| + |4.0| + |1.0| = 3.0 + 4.0 + 2.0 + 1.0 + 7.0 + 4.0 + 7.0 + 4.0 + 1.0 = 33.0 mm