A particle starts from the origin at t = 0 with an initial velocity of 5.4m/s along the positive x axis.If the acceleration is (-3.4 i + 4.9 j )m/s^2, determine (a)the velocity and (b)position of the particle at the moment it reaches its maximum x coordinate.
v(t)=vi+a*t is i the axis?
p(t)=po+vi(t)+1/2 a t^2
for max coordinate.
p'(t)=vi+at=0 in the x direction...
t=-vi/a= -5.4i/(-3.4i)=5.4/3.4) seconds
p(that time)=vi*t + 1/2 a t^2 you know vi, t, and a.
the velocity at that time?
v=vi+at
To find the velocity and position of the particle at the moment it reaches its maximum x coordinate, we need to analyze the motion of the particle.
Let's break down the problem step by step:
Step 1: Find the equation of motion in terms of time.
The acceleration is given as (-3.4i + 4.9j) m/s^2. Since the particle is moving only in the x direction, we can ignore the y-component. Therefore, the equation of motion in the x direction can be written as:
a(t) = -3.4 m/s^2
Step 2: Integrate the acceleration to find the velocity.
To find the velocity, we need to integrate the equation of motion in the x direction with respect to time (t). Since the initial velocity is given as 5.4 m/s, we can write:
v(t) = ∫a(t)dt + initial velocity
v(t) = ∫(-3.4)dt + 5.4
Integrating -3.4 with respect to t gives:
v(t) = -3.4t + 5.4
Step 3: Find the time (t) when the velocity becomes zero.
The maximum x-coordinate will occur when the velocity becomes zero. Setting v(t) = 0, we can solve for t:
-3.4t + 5.4 = 0
3.4t = 5.4
t = 5.4 / 3.4
Step 4: Substitute the time (t) back into the velocity equation to find the velocity at the maximum x-coordinate.
Substituting the value of t back into the velocity equation:
v(t) = -3.4t + 5.4
v(t) = -3.4 * (5.4 / 3.4) + 5.4
v(t) = -5.4 + 5.4
v(t) = 0
So, at the moment the particle reaches its maximum x-coordinate, its velocity is 0 m/s.
Step 5: Find the position of the particle at the maximum x-coordinate.
To find the position, we need to integrate the velocity equation with respect to time (t).
Since the initial position is at the origin, the position equation can be written as:
x(t) = ∫v(t)dt + initial position
x(t) = ∫(0)dt + 0
x(t) = 0
Therefore, at the moment the particle reaches its maximum x-coordinate, its position is also 0.
In summary:
(a) The velocity of the particle at the moment it reaches its maximum x-coordinate is 0 m/s.
(b) The position of the particle at the moment it reaches its maximum x-coordinate is 0.
To determine the velocity and position of the particle at the moment it reaches its maximum x coordinate, we need to find the time at which the particle reaches this point.
Let's denote the time when the particle reaches its maximum x coordinate as "t_max".
We can find the time when the particle reaches its maximum x coordinate using the equation for displacement in x-direction:
x = ut + (1/2)at^2
Since the particle starts at the origin, its initial position is x = 0.
Substituting the given values, we have:
0 = (5.4 m/s)(t_max) + (1/2)(-3.4 m/s^2)(t_max)^2
Simplifying this equation, we get:
(1/2)(-3.4 m/s^2)(t_max)^2 + (5.4 m/s)(t_max) = 0
Multiplying through by 2 to remove the fraction:
-3.4(t_max)^2 + 10.8(t_max) = 0
Simplifying further, we have:
-3.4(t_max)^2 + 10.8(t_max) = 0
Factorizing the equation, we get:
t_max (-3.4t_max + 10.8) = 0
So, either t_max = 0 or -3.4t_max + 10.8 = 0
Solving the second equation, we find:
-3.4t_max + 10.8 = 0
-3.4t_max = -10.8
t_max = -10.8 / -3.4
t_max ≈ 3.18 seconds
Now that we know the time at which the particle reaches its maximum x coordinate, we can determine the velocity and position of the particle at that moment.
(a) Velocity at t_max:
Using the equation for velocity:
v = u + at
Substituting the given values, we have:
v = 5.4 m/s + (-3.4 m/s^2)(3.18 s)
v ≈ 5.4 m/s - 10.892 m/s
v ≈ -5.492 m/s
So, the velocity of the particle at the moment it reaches its maximum x coordinate is approximately -5.492 m/s in the negative x-direction.
(b) Position at t_max:
Using the equation for position:
x = ut + (1/2)at^2
Substituting the given values, we have:
x = (5.4 m/s)(3.18 s) + (1/2)(-3.4 m/s^2)(3.18 s)^2
x ≈ 17.172 m + (1/2)(-3.4 m/s^2)(10.152 m^2)
x ≈ 17.172 m - 173.54 m^2/s^2
x ≈ -156.368 m
So, the position of the particle at the moment it reaches its maximum x coordinate is approximately -156.368 m.