Answer :
Let's solve this problem step-by-step.
1. Initial Conditions and Given Values:
- Initial position: [tex]\((x_0, y_0) = (0, 0)\)[/tex]
- Initial velocity: [tex]\(\vec{v_0} = 5 \hat{i} \, \text{m/s} + 0 \hat{j} \, \text{m/s}\)[/tex]
- Acceleration: [tex]\(\vec{a} = 3 \hat{i} \, \text{m/s}^2 + 2 \hat{j} \, \text{m/s}^2\)[/tex]
- We need to find the [tex]\(y\)[/tex]-coordinate when the [tex]\(x\)[/tex]-coordinate is [tex]\(84 \, \text{m}\)[/tex].
2. Equation of Motion in the [tex]\(x\)[/tex]-direction:
The equation for the [tex]\(x\)[/tex]-coordinate is given by:
[tex]\[ x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \][/tex]
Substituting the given values:
[tex]\[ 84 = 0 + 5t + \frac{1}{2} \cdot 3 t^2 \][/tex]
Simplifying, this becomes:
[tex]\[ 84 = 5t + 1.5t^2 \][/tex]
Rearrange the quadratic equation to standard form:
[tex]\[ 1.5t^2 + 5t - 84 = 0 \][/tex]
3. Solve the Quadratic Equation for [tex]\(t\)[/tex]:
The quadratic equation [tex]\(1.5 t^2 + 5 t - 84 = 0\)[/tex] can be solved using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\(a = 1.5\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = -84\)[/tex].
Plugging in these values:
[tex]\[ t = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1.5 \cdot (-84)}}{2 \cdot 1.5} \][/tex]
Simplifying inside the square root:
[tex]\[ t = \frac{-5 \pm \sqrt{25 + 504}}{3} \][/tex]
[tex]\[ t = \frac{-5 \pm \sqrt{529}}{3} \][/tex]
[tex]\[ t = \frac{-5 \pm 23}{3} \][/tex]
This yields two solutions:
[tex]\[ t_1 = \frac{18}{3} = 6 \, \text{seconds} \][/tex]
[tex]\[ t_2 = \frac{-28}{3} \approx -9.33\, \text{seconds (not physically meaningful)} \][/tex]
We use the positive solution:
[tex]\[ t = 6 \,\text{seconds} \][/tex]
4. Equation of Motion in the [tex]\(y\)[/tex]-direction:
Now that we have the time [tex]\(t = 6 \, \text{seconds}\)[/tex], we use the equation for the [tex]\(y\)[/tex]-coordinate:
[tex]\[ y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \][/tex]
Given [tex]\(y_0 = 0\)[/tex], [tex]\(v_{0y} = 0\)[/tex], and [tex]\(a_y = 2 \, \text{m/s}^2\)[/tex], we substitute these values into the equation:
[tex]\[ y = 0 + 0 + \frac{1}{2} \cdot 2 \cdot (6)^2 \][/tex]
Simplifying:
[tex]\[ y = \frac{1}{2} \cdot 2 \cdot 36 \][/tex]
[tex]\[ y = 36 \, \text{meters} \][/tex]
Therefore, the [tex]\(y\)[/tex]-coordinate of the particle at the instant when its [tex]\(x\)[/tex]-coordinate is [tex]\(84\)[/tex] meters is [tex]\(36\)[/tex] meters, not [tex]\(48\)[/tex] meters.
The correct [tex]\(y\)[/tex]-coordinate is 36 meters.
1. Initial Conditions and Given Values:
- Initial position: [tex]\((x_0, y_0) = (0, 0)\)[/tex]
- Initial velocity: [tex]\(\vec{v_0} = 5 \hat{i} \, \text{m/s} + 0 \hat{j} \, \text{m/s}\)[/tex]
- Acceleration: [tex]\(\vec{a} = 3 \hat{i} \, \text{m/s}^2 + 2 \hat{j} \, \text{m/s}^2\)[/tex]
- We need to find the [tex]\(y\)[/tex]-coordinate when the [tex]\(x\)[/tex]-coordinate is [tex]\(84 \, \text{m}\)[/tex].
2. Equation of Motion in the [tex]\(x\)[/tex]-direction:
The equation for the [tex]\(x\)[/tex]-coordinate is given by:
[tex]\[ x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \][/tex]
Substituting the given values:
[tex]\[ 84 = 0 + 5t + \frac{1}{2} \cdot 3 t^2 \][/tex]
Simplifying, this becomes:
[tex]\[ 84 = 5t + 1.5t^2 \][/tex]
Rearrange the quadratic equation to standard form:
[tex]\[ 1.5t^2 + 5t - 84 = 0 \][/tex]
3. Solve the Quadratic Equation for [tex]\(t\)[/tex]:
The quadratic equation [tex]\(1.5 t^2 + 5 t - 84 = 0\)[/tex] can be solved using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\(a = 1.5\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = -84\)[/tex].
Plugging in these values:
[tex]\[ t = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1.5 \cdot (-84)}}{2 \cdot 1.5} \][/tex]
Simplifying inside the square root:
[tex]\[ t = \frac{-5 \pm \sqrt{25 + 504}}{3} \][/tex]
[tex]\[ t = \frac{-5 \pm \sqrt{529}}{3} \][/tex]
[tex]\[ t = \frac{-5 \pm 23}{3} \][/tex]
This yields two solutions:
[tex]\[ t_1 = \frac{18}{3} = 6 \, \text{seconds} \][/tex]
[tex]\[ t_2 = \frac{-28}{3} \approx -9.33\, \text{seconds (not physically meaningful)} \][/tex]
We use the positive solution:
[tex]\[ t = 6 \,\text{seconds} \][/tex]
4. Equation of Motion in the [tex]\(y\)[/tex]-direction:
Now that we have the time [tex]\(t = 6 \, \text{seconds}\)[/tex], we use the equation for the [tex]\(y\)[/tex]-coordinate:
[tex]\[ y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \][/tex]
Given [tex]\(y_0 = 0\)[/tex], [tex]\(v_{0y} = 0\)[/tex], and [tex]\(a_y = 2 \, \text{m/s}^2\)[/tex], we substitute these values into the equation:
[tex]\[ y = 0 + 0 + \frac{1}{2} \cdot 2 \cdot (6)^2 \][/tex]
Simplifying:
[tex]\[ y = \frac{1}{2} \cdot 2 \cdot 36 \][/tex]
[tex]\[ y = 36 \, \text{meters} \][/tex]
Therefore, the [tex]\(y\)[/tex]-coordinate of the particle at the instant when its [tex]\(x\)[/tex]-coordinate is [tex]\(84\)[/tex] meters is [tex]\(36\)[/tex] meters, not [tex]\(48\)[/tex] meters.
The correct [tex]\(y\)[/tex]-coordinate is 36 meters.
