A particle travels to the right at a constant
rate of 6.1 m/s. It suddenly is given a vertical
acceleration of 1.8 m/s
2
for 4.5 s.
What is its direction of travel after the
acceleration with respect to the horizontal?
Answer between −180◦
and +180◦
.
Answer in units of ◦
What is the speed at this time?
Answer in units of m/s
1. V = a*t = 1.8m/s^2 * 4.5s. = 8.1 m/s.
X = 6.1 m/s.
Y = 8.1 m/s.
tan A = Y/X = 8.1/6.1 = 1.32787
A = 53o = Direction.
2. V = Y/sinA = 8.1/sin53 = 10.14 m/s.
To find the direction of travel after the acceleration with respect to the horizontal, we need to determine the angle θ between the particle's velocity vector before and after the vertical acceleration is applied.
Given:
Horizontal velocity (v_x) = 6.1 m/s
Vertical acceleration (a_y) = 1.8 m/s^2
Time of acceleration (t) = 4.5 s
Step 1: Determine the change in vertical velocity during the acceleration.
Using the formula: Δv_y = a_y * t
Δv_y = 1.8 m/s^2 * 4.5 s
Δv_y = 8.1 m/s
Step 2: Determine the horizontal displacement during the acceleration.
Using the formula: Δx = v_x * t
Δx = 6.1 m/s * 4.5 s
Δx = 27.45 m
Step 3: Calculate the angle θ using the tangent function.
Using the formula: θ = arctan(Δv_y / Δx)
θ = arctan(8.1 m/s / 27.45 m)
θ ≈ 16.92°
Therefore, the direction of travel after the acceleration with respect to the horizontal is approximately 16.92°.
To find the speed at this time, we can calculate the resultant velocity using the Pythagorean theorem.
Step 1: Determine the magnitude of the resultant velocity.
Using the formula: v = √(v_x^2 + v_y^2)
v = √((6.1 m/s)^2 + (8.1 m/s)^2)
v ≈ √(37.21 + 65.61)
v ≈ √102.82
v ≈ 10.14 m/s
Therefore, the speed at this time is approximately 10.14 m/s.
To determine the direction of travel after the acceleration, we need to consider the vectors involved. We have a horizontal vector representing the initial velocity and a vertical vector representing the acceleration.
Given:
Initial horizontal velocity (Vx) = 6.1 m/s
Vertical acceleration (ay) = 1.8 m/s^2
Duration of acceleration (t) = 4.5 s
First, we need to find the change in velocity in the vertical direction. We use the equation of motion: ΔV = ay * t.
ΔV = 1.8 m/s^2 * 4.5 s = 8.1 m/s
The particle gained a vertical velocity of 8.1 m/s in the upward direction. Now, we have to find the angle between the initial velocity and the final velocity using trigonometry.
tan(θ) = vertical velocity / horizontal velocity
θ = arctan(ΔV / Vx)
θ = arctan(8.1 / 6.1)
θ ≈ 54.7° (rounded to one decimal place)
However, this angle doesn't account for the direction. Since the particle was initially traveling to the right, we should consider this as a positive angle. Therefore, the direction of travel after the acceleration with respect to the horizontal is +54.7°.
To find the speed at this time, we can use the Pythagorean theorem as follows:
Speed = sqrt((horizontal velocity)^2 + (vertical velocity)^2)
Speed = sqrt((6.1 m/s)^2 + (8.1 m/s)^2)
Speed = sqrt(37.21 m^2/s^2 + 65.61 m^2/s^2)
Speed ≈ 9.64 m/s (rounded to two decimal places)
Therefore, the speed at this time is approximately 9.64 m/s.