A particle leaves the origin with an initial velocity = (2.35 m/s) and moves with constant acceleration = (-1.90 m/s2) + (2.60 m/s2)>.
(a) How far does the particle move in the x direction before turning around?
m
(b) What is the particle's velocity at this time?
( m/s) + ( m/s)
(c) Determine the particle's position x at t = 0.500 s, 1.00 s, 1.50 s, and 2.00 s.
x(0.500 s) = ( m) + ( m)
x(1.00 s) = ( m) + ( m)
x(1.50 s) = ( m) + ( m)
x(2.00 s) = ( m) + ( m)
Plot these points, and use them to sketch position versus time for the particle.
To solve this problem, we can use the following kinematic equations:
1. Position in x-direction (x) at time (t) is given by:
x = x₀ + v₀t + (1/2)at²
2. Velocity in x-direction (v) at time (t) is given by:
v = v₀ + at
Let's solve the problem step by step:
(a) To find how far the particle moves in the x-direction before turning around, we need to determine the time when the particle changes direction. This occurs when the velocity becomes zero.
Using the equation v = v₀ + at, we can set v = 0 and solve for t:
0 = v₀ + at
t = -v₀/a
Substituting the given values:
t = -2.35 m/s / (2.60 m/s² - 1.90 m/s²)
Simplifying the expression:
t ≈ -2.35 m/s / 0.70 m/s²
t ≈ -3.36 s
Since time cannot be negative in physical terms, the particle turns around at t ≈ 3.36 s. So, it is moving in the positive x-direction for 3.36 seconds before changing direction.
Now, substitute the turning point time (3.36 s) into the position equation to find the distance traveled:
x = x₀ + v₀t + (1/2)at²
For simplicity, since it starts at the origin, x₀ = 0. Hence:
x = v₀t + (1/2)at²
x = (2.35 m/s)(3.36 s) + (1/2)(-1.90 m/s² + 2.60 m/s²)(3.36 s)²
Evaluating the equation:
x ≈ 2.35 m/s * 3.36 s + (1/2)(0.70 m/s²)(11.30 s²)
x ≈ 7.89 m - 4.94 m
x ≈ 2.95 m
So, the particle moves approximately 2.95 meters in the x-direction before turning around.
(b) To find the particle's velocity at the turning point, we can directly substitute t = 3.36 s into the velocity equation:
v = v₀ + at
Substituting the values:
v = 2.35 m/s + (-1.90 m/s² + 2.60 m/s²)(3.36 s)
Evaluating the equation:
v = 2.35 m/s + 0.70 m/s² * 3.36 s
v ≈ 2.35 m/s + 2.35 m/s
v ≈ 4.70 m/s
So, the particle's velocity at the turning point is approximately 4.70 m/s in the positive x-direction.
(c) To find the particle's position at various times, substitute each given time value into the position equation:
x = x₀ + v₀t + (1/2)at²
Given values: t = 0.500 s, 1.00 s, 1.50 s, 2.00 s
For each value of t, substitute the values and calculate x.
For example, to find x at t = 0.500 s:
x(0.500 s) = 0 + (2.35 m/s)(0.500 s) + (1/2)(-1.90 m/s² + 2.60 m/s²)(0.500 s)²
Similarly, substitute other values of t (1.00 s, 1.50 s, 2.00 s) into the equation and calculate x for each.
Plotting these points on a graph and connecting them will give you the position versus time graph for the particle.