A particle leaves the origin with an initial velocity v=(2.40 m/s)x hat and moves with a constant acceleration a=(-1.90 m/s ^2)x hat+(3.20 m/s ^2)y hat. How far does the particle move in the x direction before turning around? Answer is 1.48 m. What is the particle's velocity at this time? I can't figure this second part out.
When a particle is turning around in some direction (x in this case), its velocity compnent in that direction is zero.
To find the distance the particle moves in the x-direction before turning around, we need to find the time it takes for the particle to reach the point of turning. We can use the formula:
s = ut + 0.5at^2
where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
In the x-direction, the acceleration is given as a_x = -1.90 m/s^2.
The initial velocity in the x-direction is given as v_x = 2.40 m/s.
Since the particle turns around when the velocity in the x-direction becomes zero, we can set up the equation:
0 = v_x + a_x*t
0 = 2.40 - 1.90*t
Solving this equation for t:
1.90*t = 2.40
t = 2.40 / 1.90
t ≈ 1.26 s
Now, we can find the distance traveled in the x-direction using the equation:
s_x = v_x*t + 0.5*a_x*t^2
s_x = 2.40 * 1.26 + 0.5 * (-1.90) * (1.26)^2
s_x = 3.024 - 1.1985
s_x ≈ 1.83 m
Therefore, the particle moves approximately 1.83 meters in the x-direction before turning around.
To find the particle's velocity at this time, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, we already have the initial velocity in the x-direction, which is given as v_x = 2.40 m/s. The acceleration in the x-direction is a_x = -1.90 m/s^2.
Using the same time value t = 1.26 s that we found earlier, we can calculate the final velocity in the x-direction:
v_x = 2.40 - 1.90 * 1.26
v_x ≈ -0.408 m/s
The negative sign indicates that the particle is moving in the negative x-direction (opposite to the initial direction of motion) at this time.
Therefore, the particle's velocity at the time it turns around is approximately -0.408 m/s in the x-direction.
To find the distance the particle moves in the x direction before turning around, we need to determine the time it takes for the particle to reach the point where it turns around.
First, let's find the time taken for the particle to reach the turning point.
Using the equation of motion:
x = x0 + v₀t + (1/2)at²
Since the particle starts from the origin, x₀ = 0. Therefore, the equation simplifies to:
x = (1/2)at²
For motion in the x direction, the acceleration is given by -1.90 m/s². Substituting this into the equation, we get:
x = (1/2)(-1.90 m/s²)t²
Now, let's find the time it takes to reach the turning point. At this point, the particle's velocity in the x-direction becomes zero.
Using the formula for velocity:
v = v₀ + at
Substituting the values, we can find the time taken to reach the turning point:
0 = (2.40 m/s) + (-1.90 m/s²)t_turn
Solving this equation for t_turn, we find t_turn = 1.26 seconds.
Now that we know the time taken to reach the turning point, we can find the distance traveled in the x direction.
Using the equation of motion:
x = x0 + v₀t + (1/2)at²
Setting x = 0 (since the particle reaches the turning point here) and substituting the known values, we have:
0 = 0 + (2.40 m/s)(1.26 s) + (1/2)(-1.90 m/s²)(1.26 s)² + 0
Solving this equation for x, we find x ≈ 1.48 m.
So, the particle moves approximately 1.48 m in the x-direction before turning around.
To find the particle's velocity at this time, we can use the equation:
v = v₀ + at
Substituting the values, we have:
v = (2.40 m/s) + (-1.90 m/s²)(1.26 s)
Evaluating this expression, we find the particle's velocity at the time of turning around to be approximately -0.618 m/s in the x direction.