Initially (at time t = 0) a particle is moving vertically at 5.8 m/s and horizontally at 0 m/s. Its horizontal acceleration is 2.1 m/s2.
At what time will the particle be traveling at 54� with respect to the horizontal? The acceleration due to gravity is 9.8 m/s2.
Answer in units of s
kýsadevre sorularý
To determine the time at which the particle will be traveling at an angle of 54 degrees with respect to the horizontal, we can follow these steps:
Step 1: Determine the horizontal component of velocity.
Initially, the particle is moving horizontally at 0 m/s, so its horizontal component of velocity remains constant throughout. The horizontal component of velocity is given by the formula: Vx = initial velocity = 0 m/s.
Step 2: Determine the vertical component of velocity.
At t = 0, the particle is moving vertically at a velocity of 5.8 m/s. Due to the acceleration due to gravity, the vertical component of velocity will increase at a rate of 9.8 m/s². We can use the formula: Vy = initial velocity + (gravity * time).
Step 3: Determine the total velocity.
The total velocity at any given time can be calculated using the Pythagorean theorem: V = sqrt(Vx² + Vy²).
Step 4: Calculate the time.
The objective is to find the time when the particle is traveling at an angle of 54 degrees, which means the total velocity equals 54 m/s. We'll set up an equation using the formula from Step 3 and solve for time.
Here's the equation for Step 4:
54 = sqrt((0)² + (5.8 + 9.8 * t)²)
Now, we can solve the equation to find the value of time (t) at which the particle is traveling at 54 degrees with respect to the horizontal.
To determine the time at which the particle will be traveling at an angle of 54 degrees with respect to the horizontal, we can analyze the vertical and horizontal components of its motion separately.
Given:
Initial vertical velocity (Vy0) = 5.8 m/s
Initial horizontal velocity (Vx0) = 0 m/s
Horizontal acceleration (Ax) = 2.1 m/s^2
Vertical acceleration due to gravity (Ay) = 9.8 m/s^2
Angle with respect to the horizontal (θ) = 54 degrees
First, we need to find the time at which the particle reaches a velocity of zero in the vertical direction (Vy = 0 m/s), which occurs at the maximum height of its trajectory. We can use the following kinematic equation:
Vy = Vy0 + Ay * t
0 = 5.8 m/s + (-9.8 m/s^2) * t
Solving for t:
-5.8 m/s = (-9.8 m/s^2) * t
t = -5.8 m/s / (-9.8 m/s^2)
t ≈ 0.5918 s
Next, we can calculate the horizontal distance covered by the particle until it reaches that maximum height. Since the horizontal velocity (Vx) remains constant at 0 m/s, the horizontal distance (Dx) can be calculated using the formula:
Dx = Vx0 * t + (1/2) * Ax * t^2
Dx = 0 m/s * 0.5918 s + (1/2) * 2.1 m/s^2 * (0.5918 s)^2
Dx = 0 + 0.616567 m
Dx ≈ 0.617 m
Now, we can determine the total horizontal distance covered by the particle when it reaches an angle of 54 degrees. Since the angle with respect to the horizontal is 54 degrees, we will have a right triangle formed by the vertical displacement at the maximum height and the horizontal displacement.
Using trigonometry, we can calculate the total horizontal distance (D) covered by the particle using the following relation:
tan θ = Dx / D
tan 54° = 0.617 m / D
D ≈ 0.617 m / tan 54°
D ≈ 0.617 m / 1.37638192047
D ≈ 0.448 m
Finally, we need to find the time taken to cover this horizontal distance, D. We can use the formula:
D = Vx0 * t + (1/2) * Ax * t^2
Rearranging the equation:
(1/2) * Ax * t^2 + Vx0 * t - D = 0
Solving this quadratic equation for t:
t = (-Vx0 ± √(Vx0^2 - 4 * (1/2) * Ax * (-D))) / (2 * (1/2) * Ax)
t = (-0 ± √(0^2 - 4 * 0.5 * 2.1 * (-0.448))) / (2 * 0.5 * 2.1)
t = (√(0.471264) / 2.1
t ≈ 0.315 s
Therefore, the particle will be traveling at 54 degrees with respect to the horizontal at approximately t = 0.315 seconds.