A particle starts from the origin at `t= 0` with a velocity of `8.0 hat j m//s` and moves in the x-y plane with a constant acceleration of `(4.0 hat i + 2.0 hat j) m//s^2.` At the instant the particle's x-coordinate is 29 m, what are
(a) its y-coordinate and
(b) its speed ?

Explanation

To solve the problem step by step, we will break it down into two parts: finding the y-coordinate when the x-coordinate is 29 m, and then calculating the speed of the particle at that instant.

### Step 1: Find the time when the x-coordinate is 29 m

The equation of motion in the x-direction is given by:
\[ x = u_x t + \frac{1}{2} a_x t^2 \]
where:
- \(u_x = 0\) (initial velocity in the x-direction is 0 m/s),
- \(a_x = 4.0 \, \hat{i} \, \text{m/s}^2\) (acceleration in the x-direction),
- \(x = 29 \, \text{m}\).

Substituting the values, we have:
\[ 29 = 0 + \frac{1}{2} (4) t^2 \]
This simplifies to:
\[ 29 = 2t^2 \]
\[ t^2 = \frac{29}{2} \]
\[ t = \sqrt{14.5} \approx 3.8 \, \text{s} \]

### Step 2: Find the y-coordinate at \(t = 3.8 \, \text{s}\)

The equation of motion in the y-direction is given by:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
where:
- \(u_y = 8.0 \, \hat{j} \, \text{m/s}\) (initial velocity in the y-direction),
- \(a_y = 2.0 \, \hat{j} \, \text{m/s}^2\) (acceleration in the y-direction).

Substituting the values, we have:
\[ y = 8 \cdot 3.8 + \frac{1}{2} (2) (3.8)^2 \]
Calculating each term:
\[ y = 30.4 + \frac{1}{2} (2) (14.44) \]
\[ y = 30.4 + 14.44 \]
\[ y \approx 44.84 \, \text{m} \approx 45 \, \text{m} \]

### Step 3: Find the speed of the particle at \(t = 3.8 \, \text{s}\)

The velocity in the x-direction at time \(t\) is given by:
\[ v_x = u_x + a_x t \]
Substituting the values:
\[ v_x = 0 + (4)(3.8) = 15.2 \, \text{m/s} \]

The velocity in the y-direction at time \(t\) is given by:
\[ v_y = u_y + a_y t \]
Substituting the values:
\[ v_y = 8 + (2)(3.8) = 8 + 7.6 = 15.6 \, \text{m/s} \]

### Step 4: Calculate the net speed

The net speed \(v\) is given by:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(15.2)^2 + (15.6)^2} \]
Calculating:
\[ v = \sqrt{231.04 + 243.36} = \sqrt{474.4} \approx 21.77 \, \text{m/s} \approx 22 \, \text{m/s} \]

### Final Answers:
(a) The y-coordinate is approximately **45 m**.
(b) The speed of the particle is approximately **22 m/s**.