Q.A particle starts from origin at $t = 0$ with a velocity $5.0\,\hat{i}\ \text{m/s}$ and moves in $x$-$y$ plane under action of a force which produces a constant acceleration of $(3.0\,\hat{i} + 2.0\,\hat{j})\ \text{m/s}^{2}$.

The key idea is to apply the kinematic equations for motion under constant acceleration in vector form, resolving them into components as needed.

1. The position vector $\vec{r}(t)$ is given by $\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2}\vec{a}t^2$. Substituting the given values:

   $\vec{r}(t) = \vec{0} + (5.0\,\hat{i})t + \frac{1}{2}(3.0\,\hat{i} + 2.0\,\hat{j})t^2$

   $\vec{r}(t) = (5.0t + 1.5t^2)\hat{i} + (1.0t^2)\hat{j}$.

   Thus, the $x$-coordinate is $x(t) = 5.0t + 1.5t^2$ and the $y$-coordinate is $y(t) = 1.0t^2$.

2. To find the time when $x = 84\ \text{m}$, we solve $1.5t^2 + 5.0t - 84 = 0$. Multiplying by 2 gives $3t^2 + 10t - 168 = 0$. Using the quadratic formula $t = \frac{-10 \pm \sqrt{10^2 - 4(3)(-168)}}{2(3)} = \frac{-10 \pm \sqrt{100 + 2016}}{6} = \frac{-10 \pm \sqrt{2116}}{6} = \frac{-10 \pm 46}{6}$. Since $t > 0$, we take $t = \frac{-10 + 46}{6} = 6\ \text{s}$.

3. (a) The $y$-coordinate at $t = 6\ \text{s}$ is $y(6) = 1.0(6)^2 = 36\ \text{m}$.

4. (b) The velocity vector $\vec{v}(t)$ is given by $\vec{v}(t) = \vec{v}_0 + \vec{a}t$. Substituting values: …

When a particle moves in two or three dimensions under constant acceleration, its motion can be analyzed by treating each spatial dimension (like $x$ and $y$) independently. This is because the acceleration in one direction does not affect the motion in a perpendicular direction. We can apply the familiar one-dimensional kinematic equations to the $x$-component of motion and the $y$-component of motion separately. Once we have the component-wise descriptions, we can combine them to find the overall position, velocity, or speed.

Here, we are given initial velocity and constant acceleration as vectors. We will first break these vectors into their $x$ and $y$ components. Then, we will use the position equation for the $x$-component to find the time at which the $x$-coordinate reaches $84\,\text{m}$. With this time, we can find the corresponding $y$-coordinate and the components of velocity, which will allow us to calculate the speed.

1. Identify Initial Conditions and Acceleration Components:

   The particle starts from the origin at $t=0$, so its initial position vector is $\vec{r}_0 = 0\,\hat{i} + 0\,\hat{j}$.

   The initial velocity is given as $\vec{v}_0 = 5.0\,\hat{i}\,\text{m/s}$.

   This means the initial $x$-component of velocity is $v_{0x} = 5.0\,\text{m/s}$, and the initial $y$-component of velocity is $v_{0y} = 0\,\text{m/s}$.

   The constant acceleration is $\vec{a} = (3.0\,\hat{i} + 2.0\,\hat{j})\,\text{m/s}^2$.

   So, the $x$-component of acceleration is $a_x = 3.0\,\text{m/s}^2$, and the $y$-component of acceleration is $a_y = 2.0\,\text{m/s}^2$.

2. Recall Kinematic Equations for Position and Velocity:

   For motion with constant acceleration, the position vector $\vec{r}(t)$ and velocity vector $\vec{v}(t)$ at any time $t$ are given by:

   $\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2}\vec{a}t^2$

   $\vec{v}(t) = \vec{v}_0 + \vec{a}t$

   We can write these equations in terms of their $x$ and $y$ components:

   For the $x$-component:

   $x(t) = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$

   $v_x(t) = v_{0x} + a_xt$

   For the $y$-component:

   $y(t) = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

   $v_y(t) = v_{0y} + a_yt$

3. Substitute Initial Values into Component Equations:

   Using the values from Step 1:

   $x_0 = 0$, $v_{0x} = 5.0$, $a_x = 3.0$

   $y_0 = 0$, $v_{0y} = 0$, $a_y = 2.0$

   The position equations become:

   $x(t) = 0 + 5.0t + \frac{1}{2}(3.0)t^2 \implies x(t) = 5.0t + 1.5t^2$

   $y(t) = 0 + 0t + \frac{1}{2}(2.0)t^2 \implies y(t) = t^2$

   The velocity equations become:

   $v_x(t) = 5.0 + 3.0t$

   $v_y(t) = 0 + 2.0t \implies v_y(t) = 2.0t$

4. Determine the Time when $x$-coordinate is $84\,\text{m}$ (Part a):

   We are given that the $x$-coordinate is $84\,\text{m}$. We use the equation for $x(t)$:

   $84 = 5.0t + 1.5t^2$

   Rearrange this into a standard quadratic equation:

   $1.5t^2 + 5.0t - 84 = 0$

   We can solve for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

   $t = \frac{-5.0 \pm \sqrt{(5.0)^2 - 4(1.5)(-84)}}{2(1.5)}$

   $t = \frac{-5.0 \pm \sqrt{25 + 504}}{3}$

   $t = \frac{-5.0 \pm \sqrt{529}}{3}$

   $t = \frac{-5.0 \pm 23}{3}$

   This gives two possible values for $t$: …

Concept: Getting Time Without Solving a Quadratic

Method: Time-Free Equation First ($v_x^2=v_{0x}^2+2a_x x$), Then a Linear Solve for $t$

The position equation $x(t)=5.0t+1.5t^2=84$ is a genuine quadratic in $t$, needing the quadratic formula. This method sidesteps that entirely: first find $v_x$ at $x=84$ directly from the time-free kinematic relation (which only involves $x$, not $t$), then get $t$ from a single linear equation $v_x=v_{0x}+a_xt$ -- no quadratic formula anywhere.

Steps

1. Identify the $x$-motion's known quantities: $v_{0x}=5.0\ \text{m/s}$, $a_x=3.0\ \text{m/s}^2$, and the target $x=84\ \text{m}$ (with $x_0=0$).

2. Apply the time-free relation along $x$ -- this equation never involves $t$, only displacement:

(Taking the positive root: the particle starts moving in $+x$ and only accelerates further in $+x$, so $v_x$ stays positive throughout.)

3. Now solve for $t$ from the LINEAR velocity equation $v_x = v_{0x}+a_xt$, instead of the quadratic position equation:

4. With $t$ now known, find the $y$-coordinate using the $y$-motion's own position equation (there is no shortcut needed here, since $v_{0y}=0$ makes it already simple):

5. Find $v_y$ at this instant:

6. Speed = magnitude of the full velocity vector: …