A particle moving initially at 4 m/s in the positive x direction is given an acceleration of 3 m/s^2 in the positive y direction for 2 seconds.
What is the magnitude and direction of the particle's velocity after 2 seconds?
If the particle was located at the origin when the acceleration began, what are its x and y coordinates after 2 seconds?
V = a * t = 3 m/s^2 * 2 s = 6 m/s in
positive y-direction,
V = 4 + i6,
tanA = Y/X = 6/4 = 1.5,
A = 56.3 Deg.,
MAG. = X/cosA = 4/cos = 7.2 m,
V = 4 + i6 = 7.2 m @ 56.3 Deg.,
(4 , 6).
To find the magnitude and direction of the particle's velocity after 2 seconds, we can use the equations of motion.
Given:
Initial velocity (u): 4 m/s in the positive x direction
Acceleration (a): 3 m/s^2 in the positive y direction
Time (t): 2 seconds
First, we need to find the final velocity (v) using the formula:
v = u + at
Substituting the given values, we have:
v = 4 m/s + 3 m/s^2 * 2 s
v = 4 m/s + 6 m/s
v = 10 m/s
So, the magnitude of the particle's velocity after 2 seconds is 10 m/s.
Next, we can find the direction of the velocity. Since the acceleration is in the positive y direction, the final velocity will have a positive y component. However, the initial velocity is in the positive x direction, which means the final velocity will also have a positive x component.
Therefore, the particle's velocity after 2 seconds can be represented as (10 m/s, 0 m/s) in vector notation. The direction of the velocity is along the positive x-axis.
To find the x and y coordinates of the particle after 2 seconds, we can use the equations of motion for displacement. Since the initial position is at the origin, the initial coordinates (x₀, y₀) are (0, 0).
The x-coordinate can be calculated using the formula:
x = x₀ + ut + (1/2)at^2
Substituting the given values, we have:
x = 0 + (4 m/s)(2 s) + (1/2)(0 m/s^2)(2 s)^2
x = 0 + 8 m + 0 m
x = 8 m
So, the x-coordinate of the particle after 2 seconds is 8 m.
The y-coordinate can be calculated using the formula:
y = y₀ + vt + (1/2)at^2
Substituting the given values, we have:
y = 0 + (0 m/s)(2 s) + (1/2)(3 m/s^2)(2 s)^2
y = 0 + 0 m + (1/2)(3 m/s^2)(4 s^2)
y = 0 + 0 m + 6 m
y = 6 m
So, the y-coordinate of the particle after 2 seconds is 6 m.
Therefore, the x and y coordinates of the particle after 2 seconds are (8 m, 6 m).