A particle moves in the xy plane with constant acceleration. At time zero, the particle is at

x = 6.0 m, y = 3.0 m, and has velocity v = 4.0 m/s + -1.0 m/s . The acceleration is given
by the vector a = 4.0 m/s2 + 0 m/s2 .

(a) Find the velocity vector at t = 3.0 s.
?( m/s) + ?( m/s)

(b) Find the position vector at t = 3.0 s.
?( m) + ?( m)

(c) Give the magnitude and direction of the position vector in part (b).
?m
?° (counterclockwise from the +x-axis)

r0 =(6i + 3j)

V0 =(4i + -1j)
a =(4i + 0j)

(a) velocity v1 at t=3 s
v1 = v0 + at
v1 = (4i + -1j) + (4i + 0j)3
v1 = (4i + -1j) + (12i + 0j)
vi = (16i + -1j)

(b) position r1 at t=3 s
r1 = r0 + v0.t + (a.t^2)/2
r1 = (36i + 0j)

(c)
Magnitude
|r1| =sqrt[(x1)^2 + (y1)^2] = 36
Angle
theta =arctan(y1/x1) = 0 degree

(a) Well, well, well, let's see what we have here. To find the velocity vector at t = 3.0 s, we can use the good ol' kinematic equations. Since the acceleration is constant, we have 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.

Given that the initial velocity v is (4.0 m/s, -1.0 m/s) and the acceleration a is (4.0 m/s^2, 0 m/s^2), we can plug these values into the equation to find the final velocity vector.

So, v = u + at
v = (4.0 m/s, -1.0 m/s) + (4.0 m/s^2, 0 m/s^2) * 3.0 s

Calculating it with a grin, we get:
v = (4.0 m/s + 4.0 m/s^2 * 3.0 s, -1.0 m/s + 0 m/s^2 * 3.0 s)

Don't worry, I won't keep you waiting. Crunching the numbers, we find:
v = (12.0 m/s, -1.0 m/s)

So, the velocity vector at t = 3.0 s is (12.0 m/s, -1.0 m/s).

(b) Now, let's move on to finding the position vector at t = 3.0 s. Again, we can use the kinematic equations, but this time we'll use the equation s = ut + (1/2)at^2, where s is the position vector.

Given that the initial position vector s is (6.0 m, 3.0 m), the initial velocity u is (4.0 m/s, -1.0 m/s), the acceleration a is (4.0 m/s^2, 0 m/s^2), and the time t is 3.0 s, we can plug these values into the equation to find the position vector.

So, s = ut + (1/2)at^2
s = (6.0 m, 3.0 m) + (4.0 m/s, -1.0 m/s) * 3.0 s + (1/2) * (4.0 m/s^2, 0 m/s^2) * (3.0 s)^2

Ready for some calculations? Here we go:
s = (6.0 m, 3.0 m) + (4.0 m/s, -1.0 m/s) * 3.0 s + (1/2) * (4.0 m/s^2, 0 m/s^2) * (9.0 s^2)

Brace yourself, the answer is coming:
s = (6.0 m + 12.0 m, 3.0 m - 3.0 m) + (2.0 m/s, 0 m/s) * (9.0 s^2)

Drumroll, please! After some math, we get:
s = (18.0 m, 0 m) + (18.0 m, 0 m)
s = (36.0 m, 0 m)

So, the position vector at t = 3.0 s is (36.0 m, 0 m).

(c) Ah, we're almost there! Now, to find the magnitude and direction of the position vector in part (b), we simply need to calculate the magnitude and angle.

The magnitude, my friend, is simply the distance from the origin to the point represented by the position vector, which we already have as (36.0 m, 0 m).

By using the Pythagorean theorem, we can calculate the magnitude:
magnitude = sqrt((36.0 m)^2 + (0 m)^2)

Gotta love those square roots, right? After doing the math, we find:
magnitude = 36.0 m

Now, for the angle! The direction is given counterclockwise from the +x-axis. Since the position vector is aligned with the +x-axis, the angle would be 0 degrees.

So, the magnitude of the position vector is 36.0 m, and the direction is 0 degrees counterclockwise from the +x-axis.

And there you have it, my friend! The answers to all your burning questions.

To find the velocity vector at t = 3.0 s, we need to use the formula:

v = v0 + at

where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.

Given:
v0 = 4.0 m/s + -1.0 m/s
a = 4.0 m/s^2 + 0 m/s^2
t = 3.0 s

(a) To find the velocity vector at t = 3.0 s, we substitute the given values into the formula:

v = (4.0 m/s + -1.0 m/s) + (4.0 m/s^2 + 0 m/s^2) * 3.0 s

v = 3.0 m/s + (4.0 m/s^2 * 3.0 s)

v = 3.0 m/s + 12.0 m/s

v = 15.0 m/s

Therefore, the velocity vector at t = 3.0 s is (15.0 m/s) in the x-direction.

(b) To find the position vector at t = 3.0 s, we need to use the formula:

Δr = v0t + 0.5at^2

where Δr is the displacement vector, v0 is the initial velocity, a is the acceleration, and t is the time.

Given:
v0 = 4.0 m/s + -1.0 m/s
a = 4.0 m/s^2 + 0 m/s^2
t = 3.0 s

Substituting the given values into the formula:

Δr = (4.0 m/s + -1.0 m/s) * 3.0 s + 0.5 * (4.0 m/s^2 + 0 m/s^2) * (3.0 s)^2

Δr = 9.0 m + 0.5 * 4.0 m/s^2 * 9.0 s^2

Δr = 9.0 m + 18.0 m

Δr = 27.0 m

Therefore, the position vector at t = 3.0 s is (27.0 m) in the x-direction.

(c) The magnitude of the position vector in part (b) is given by the formula:

|Δr| = √(x^2 + y^2)

where Δr is the position vector.

Given:
Δr = 27.0 m

The magnitude is:

|Δr| = √(27.0 m^2)

|Δr| = 27.0 m

The direction of the position vector can be found using the formula:

θ = tan⁻¹(y/x)

where Δr = (x, y).

Given:
Δr = 27.0 m in the x-direction (since the y-component is zero)

The direction is:

θ = tan⁻¹(0/27.0)

θ = tan⁻¹(0)

θ = 0° (counterclockwise from the +x-axis)

Therefore, the magnitude of the position vector is 27.0 m, and the direction is 0° counterclockwise from the +x-axis.

To solve this problem, we need to understand the equations of motion for constant acceleration:

1. Velocity equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

2. Displacement equation: s = ut + 0.5at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

Let's solve each part of the problem:

(a) Find the velocity vector at t = 3.0 s.

Given:
Initial velocity (u) = 4.0 m/s i - 1.0 m/s j
Acceleration (a) = 4.0 m/s^2 i + 0 m/s^2 j
Time (t) = 3.0 s

To find the velocity vector at t = 3.0 s, we can use the velocity equation:
v = u + at

Substituting the given values:
v = (4.0 m/s i - 1.0 m/s j) + (4.0 m/s^2 i + 0 m/s^2 j) * 3.0 s

Calculating the result:
v = (4.0 m/s i - 1.0 m/s j) + (12.0 m/s i + 0 m/s j)
= (16.0 m/s i - 1.0 m/s j)

The velocity vector at t = 3.0 s is (16.0 m/s i - 1.0 m/s j).

(b) Find the position vector at t = 3.0 s.

We can find the position vector using the displacement equation:
s = ut + 0.5at^2

Substituting the known values:
s = (4.0 m/s i - 1.0 m/s j) * 3.0 s + 0.5 * (4.0 m/s^2 i + 0 m/s^2 j) * (3.0 s)^2

Calculating the result:
s = (12.0 m i - 3.0 m j) + 0.5 * (12.0 m i + 0 m j) * 9.0 s^2
= (12.0 m i - 3.0 m j) + (54.0 m i + 0 m j)
= (66.0 m i - 3.0 m j)

The position vector at t = 3.0 s is (66.0 m i - 3.0 m j).

(c) Give the magnitude and direction of the position vector in part (b).

To find the magnitude of the position vector, we use the formula:
Magnitude = √(x^2 + y^2)

Substituting the values:
Magnitude = √(66.0^2 + (-3.0)^2)
= √(4356.0 + 9.0)
= √4365.0
≈ 66.08 m

To find the direction, we can use trigonometry:
θ = tan^(-1)(y / x)

Substituting the values:
θ = tan^(-1)(-3.0 / 66.0)
≈ -2.6°

The magnitude of the position vector is approximately 66.08 m, and its direction is approximately -2.6° (counterclockwise from the +x-axis).