Final answer:
The particle's final speed after experiencing constant acceleration for 3.65 seconds is approximately 1.53 m/s.
Step-by-step explanation:
The question asks for the speed of a particle at the end of a certain time interval, during which it experiences constant acceleration. Initial velocity vector of the particle is (-1.34î + 1.14è) m/s, and it accelerates at -0.11è m/s² for 3.65 seconds. To find the final speed, we first determine the final velocity components.
Initial velocity components:
vx = -1.34 m/s (x-component)
vy0 = 1.14 m/s (initial y-component)
Acceleration:
ay = -0.11 m/s² (y-component)
Time: t = 3.65 s
Final y-velocity (vy) after time t can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is acceleration, and t is time. Thus, vy = 1.14 m/s + (-0.11 m/s²)(3.65 s) = 1.14 m/s - 0.4015 m/s = 0.7385 m/s.
The x-component of the velocity does not change since there is no acceleration in the x-direction. Therefore, vx remains -1.34 m/s.
Now, to find the final speed of the particle, we use the Pythagorean theorem to combine the x and y velocity components: speed = √(vx² + vy²). Substituting the values, we get speed = √((-1.34 m/s)² + (0.7385 m/s)²) ≈ √(1.7956 + 0.5456) m/s ≈ √(2.3412) m/s ≈ 1.53 m/s.