Saurabh Koranglekar
Last Activity: 5 Years ago
We're dealing with two-dimensional motion under **constant acceleration**. The particle starts at the origin with a velocity of 8.0 m/s purely along the x-axis and then accelerates at 1.5 m/s² in a direction inclined at 37° to the x-axis. Let's find its velocity and position as functions of time.
Given Data
- Initial position: r₀ = 0
- Initial velocity: v₀ = 8.0 m/s in x-direction → v₀ = (8.0, 0)
- Acceleration: a = 1.5 m/s² at 37° to x-axis
Step 1: Resolve Acceleration into Components
Using basic trigonometry:
- ax = a·cos(37°) = 1.5 × 0.7986 ≈ 1.198 m/s²
- ay = a·sin(37°) = 1.5 × 0.6018 ≈ 0.903 m/s²
Step 2: Find Velocity as a Function of Time
Velocity after time t is:
v(t) = v₀ + a·t = (8.0 + 1.198·t, 0 + 0.903·t) = (8.0 + 1.198t, 0.903t) m/s
Step 3: Find Position as a Function of Time
Use the equation:
r(t) = r₀ + v₀·t + (1/2)·a·t²
Compute x and y positions separately:
- x(t) = 8.0·t + ½·1.198·t² = 8.0t + 0.599t²
- y(t) = 0 + ½·0.903·t² = 0.4515t²
Final Answers
Velocity of the particle at time t:
v(t) = (8.0 + 1.198t) î + (0.903t) ĵ m/s
Position of the particle at time t:
r(t) = (8.0t + 0.599t²) î + (0.4515t²) ĵ m
If you'd like the values at a specific time, I can plug them in and compute exact numerical results too.
