A particle leaves the origin with an initial velocity of 4.90 m/s in the x direction, and moves with constant acceleration ax = -2.70 m/s2 and ay = 3.90 m/s2. How far does the particle move in the x direction before turning around?
What is the particle's velocity at this time? Enter the x component first, followed by the y component.
Why did the particle turn around in the first place? Maybe it saw a mirror and got scared of its own reflection!
But let's get back to the question. To find out how far the particle moves in the x-direction before turning around, we need to find the time it takes for it to turn around.
Using the equation of motion, we can find the time it takes for the particle to turn around:
vf = vi + at
0 = 4.90 - 2.70t
Now, let's solve for t:
2.70t = 4.90
t = 4.90 / 2.70
Calculating this, we find that t ≈ 1.81 seconds.
To find the particle's velocity at this time, we can use the equations:
vx = vix + axt
vy = viy + ayt
Substituting the given values, we have:
vx = 4.90 + (-2.70)(1.81)
vy = 0 + (3.90)(1.81)
Calculating these values, we find that vx ≈ -0.89 m/s and vy ≈ 7.06 m/s.
So, the particle's velocity at the time it turns around is approximately (-0.89 m/s, 7.06 m/s).
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 change its velocity in the x direction from positive to negative.
Using the equation for displacement in the x direction with constant acceleration, we have:
xf = xi + vix * t + (1/2) * ax * t^2
Since the particle starts at the origin (xi = 0) and has an initial velocity in the x direction (vix = 4.90 m/s), the equation simplifies to:
xf = 4.90 * t - (1/2) * 2.70 * t^2
To find the time it takes for the particle to turn around, we set the final x displacement (xf) to zero:
0 = 4.90 * t - (1/2) * 2.70 * t^2
Rearranging the equation, we have:
(1/2) * 2.70 * t^2 - 4.90 * t = 0
Simplifying further:
1.35 * t^2 - 4.90 * t = 0
Now we can solve this quadratic equation for t.
Using the quadratic formula, we have:
t = (-b ± √(b^2 - 4ac)) / 2a
where a = 1.35, b = -4.90, and c = 0.
Calculating the discriminant:
D = b^2 - 4ac
= (-4.90)^2 - 4 * 1.35 * 0
= 24.01
Since the discriminant is positive, we have two real solutions for t.
Using the quadratic formula:
t = (-(-4.90) ± √(24.01)) / (2 * 1.35)
= (4.90 ± 4.90) / 2.70
Taking the positive root:
t = (4.90 + 4.90) / 2.70
= 9.80 / 2.70
≈ 3.63 s
So, the particle takes approximately 3.63 seconds to turn around.
Now, let's calculate the particle's velocity at this time. The x component of the velocity remains constant, while the y component is affected by the constant acceleration in the y direction.
The x component of the velocity (Vx) remains 4.90 m/s throughout.
The y component of the velocity (Vy) can be calculated using the equation for velocity with constant acceleration:
Vf = Vi + a * t
Since the initial velocity in the y direction (Vi) is 0 m/s and the acceleration in the y direction (ay) is 3.90 m/s^2, the equation simplifies to:
Vf = 3.90 * t
Substituting the value of t we found earlier:
Vf = 3.90 * 3.63
≈ 14.15 m/s
Therefore, at the time the particle turns around, its velocity is approximately 4.90 m/s in the x direction and 14.15 m/s in the y direction.
To find the distance the particle moves in the x direction before turning around, we can use the equation of motion:
x = x0 + v0x * t + (1/2) * ax * t^2
where x is the distance travelled in the x direction, x0 is the initial position (which is 0 in this case), v0x is the initial velocity in the x direction, ax is the acceleration in the x direction, and t is the time.
Since the particle is turning around, its final velocity in the x direction will be 0. So we can use this information to find the time it takes to reach this point.
0 = v0x + ax * t
Solving this equation for t, we get:
t = -v0x / ax
Substituting the given values, we get:
t = -4.90 m/s / -2.70 m/s^2
Simplifying:
t = 1.815 seconds
Now we can find the distance the particle moves in the x direction using the equation of motion:
x = x0 + v0x * t + (1/2) * ax * t^2
Substituting the given values, we get:
x = 0 + 4.90 m/s * 1.815 s + (1/2) * -2.70 m/s^2 * (1.815 s)^2
Simplifying:
x = 4.90 m/s * 1.815 s - 1.200 m * (1.815 s)^2
x = 8.90 m - 3.43 m
x = 5.47 m
Therefore, the particle moves a distance of 5.47 meters in the x direction before turning around.
To find the particle's velocity at this time, we can use the equations:
v = v0 + a * t
where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.
For the x component:
vx = v0x + ax * t
Substituting the given values:
vx = 4.90 m/s + -2.70 m/s^2 * 1.815 s
Simplifying:
vx = 4.90 m/s - 4.90 m/s
vx = 0 m/s
For the y component:
vy = v0y + ay * t
Substituting the given values:
vy = 0 m/s + 3.90 m/s^2 * 1.815 s
Simplifying:
vy = 0 m/s + 7.08 m/s
vy = 7.08 m/s
Therefore, the particle's velocity at this time is 0 m/s in the x direction and 7.08 m/s in the y direction.