A particle starts at the origin at t=0 with an intial velocity of 5.0m/s along the positive x axis. If the acceleration is (-3.0,4.5) m/s^2, determine the velocity and position of the particle at the moment it reaches its maximum x coordinate. You can't use a calculator.
max distance is when at= -initial velocity
t=initial velocity/a in the x direction=
t=5/3 seconds
at that point, the only velocity is j component in the y direction,or 4.5*5/3 j
v=9/5*5/3=3m/s
position:
x componant=5*5/3-1/2 3* 25/9
y component=1/2( 9/5 *25/9
To determine 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.
We know that the particle starts at the origin with an initial velocity of 5.0 m/s along the positive x-axis. This means we can represent the initial velocity as v₀ = (5.0, 0) m/s.
The acceleration of the particle is given as a = (-3.0, 4.5) m/s².
To find the velocity at any given time t, we can use the equation:
v = v₀ + at
Since the acceleration is constant, we can integrate the equation to find the position at any given time t:
r = r₀ + v₀t + 0.5at²
To find the maximum x-coordinate, we need to determine the time when the velocity in the x-direction becomes zero.
Let's first find the time at which the velocity is zero:
0 = v₀x + axt
0 = 5.0 - 3.0t
3.0t = 5.0
t = 5.0/3.0
t = 1.67 seconds
Now, we can calculate the velocity and position of the particle at t = 1.67 seconds:
Velocity:
v = v₀ + at
v = (5.0, 0) + (-3.0, 4.5) * 1.67
v = (5.0, 0) + (-5.01, 7.5)
v = (5.0 - 5.01, 0 + 7.5) m/s
v = (-0.01, 7.5) m/s
Position:
r = r₀ + v₀t + 0.5at²
r = (0, 0) + (5.0, 0) * 1.67 + 0.5 * (-3.0, 4.5) * (1.67)²
r = (0, 0) + (8.35, 0) + 0.5 * (-5.01, 7.5) * 2.79
r = (0 + 8.35 - 7.8693, 0 + 0 - 10.4635)
r ≈ (0.4807, -10.4635) meters
Therefore, at the moment it reaches its maximum x-coordinate, the particle will have a velocity of (-0.01, 7.5) m/s and a position of approximately (0.4807, -10.4635) meters.