Question

a particle with velocity vo=-2i+4j in metres per second at t=0 undergoes a constant acceleration a of magnitude a=3m/s at an angle 130° from the positive direction of the x axis. what is the particle’s velocity v at t=5s, in unit vector notation and as a magnitude and an angle​

Answer 1

The velocity of the particle at t = 5 s is approximately -8.125i + 16.13j m/s. The magnitude of the velocity is approximately 17.86 m/s, and the angle is approximately 115.2°.

Sure, let's break down the calculations step by step.

Given data:

Initial velocity v₀ = -2i + 4j m/s

Acceleration magnitude a = 3 m/s²

Acceleration angle θ = 130°

Time t = 5 s

Components of Acceleration (ax and ay):

ax = a * cos(θ)

ay = a * sin(θ)

Substitute the values:

ax = 3 * cos(130°)

ay = 3 * sin(130°)

Calculate:

ax ≈ -1.225 m/s²

ay ≈ 2.826 m/s²

Velocity Components (vx and vy):

vx = v₀x + ax * t

vy = v₀y + ay * t

Substitute the values:

vx = -2 + (-1.225) * 5

vy = 4 + 2.826 * 5

Calculate:

vx ≈ -8.125 m/s

vy ≈ 16.13 m/s

Velocity Vector (v):

v = vx * i + vy * j

v ≈ -8.125i + 16.13j

Magnitude of Velocity (|v|):

|v| = sqrt(vx² + vy²)

Calculate:

|v| ≈ 17.86 m/s

Angle of Velocity (θv):

θv = atan(vy / vx)

Calculate:

θv ≈ 115.2°