91Ó°ÊÓ

When dealing with two-dimensional motion, constant acceleration is a key concept. This means that the acceleration of the particle does not change over time. In this problem, the acceleration vector \(\vec{a} = \(4.0 \mathrm{~m} / \mathrm{s}^{2}\) \hat{\mathrm{i}}+ \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) \hat{\mathrm{j}}\) has two components: one in the x-direction and one in the y-direction.

The constant acceleration tells us that the particle's speed and direction of motion will change at a constant rate. Algebraic manipulation of the kinematic equations allows us to predict the particle's future position and velocity.

Specifically, in constant acceleration problems, we often use these equations:

• \(a = constant\)
• \(x = v_{0x}t + \frac{1}{2}a_xt^2\)
• \(y = v_{0y}t + \frac{1}{2}a_yt^2\)

By understanding these equations, you can determine not only where the particle will be, but also how fast it will be moving at any point in time.

In two-dimensional kinematics, velocity is also broken down into components. We use \(\hat{\mathrm{i}}\) to represent the x-component and \(\hat{\mathrm{j}}\) for the y-component.

For instance, the initial velocity given in this problem is \((8.0 \mathrm{~m}/\mathrm{s}) \-hat{\mathrm{j}}\). This means the particle starts with a speed of 8.0 m/s in the y-direction and no initial velocity in the x-direction. When acceleration is constant, velocity components change linearly with time.

These are the main equations to update velocity components:

• \(v_x = v_\{0x} + a_x t\)
• \(v_y = v_\{0y} + a_y t\)

This problem requires finding the particle's speed when \( x \) is 29 m. With these equations, we first calculate each velocity component at that instant and later determine the resultant speed.

Knowing all this is essential for disentangling the x and y contributions to the particle's motion.

Kinematic equations are the backbone of solving two-dimensional motion problems. These equations connect displacement, velocity, acceleration, and time.

In this exercise, we use the kinematic equation for x to find the time when x is 29 m:

\[ x = v_\{0x}t + \frac{1}{2}a_xt^2\]

Given that the initial x-velocity \(v_\{0x}\) is zero, the equation simplifies to:

\[ 29 = \frac{1}{2} \cdot 4 \cdot t^2\]

Solving this finds \( t \approx 3.81 s \). We then use this time to find the y-coordinate using a similar kinematic equation for y.

The final position and velocity equations for y are:

• \(y = v_\{0y} t + \frac{1}{2} a_y t^2 \)
• \(v_y = v_\{0y} + a_y t\)

Once we have the velocity components \(v_x\) and \(v_y\), combining them using the Pythagorean Theorem gives the resultant speed:

• \(v = \sqrt{v_x^2 + v_y^2}\)

Understanding and applying these equations is crucial for effectively solving kinematics problems in two dimensions.