Motion In A Plane With Constant Acceleration

Suppose that an object is moving in x-y plane and its acceleration a is constant. Over an interval of time, the average acceleration will equal this constant value. Now, let the velocity of the object be $\mathbf{v_0}$ at time t = 0 and v at time t. Then, by definition

$\mathbf{a}=\frac{\mathbf{v} -\mathbf{v_0}}{t-0}$

Or, $\mathbf{v}=\mathbf{v_0}+\mathbf{a}t$

In terms of components:

v$_x$ = v$_{0x}$ + a$_{x}$t

v$_y$ = v$_{0y}$ + a$_{y}$t

Let us now find how the position r changes with time. We follow the method used in the one dimensional case. Let r$_0$and r be the position vectors of the particle at time 0 and t and let the velocities at these instants be v$_0$ and v. Then, over this time interval t, the average velocity  is $(\mathbf{v_0}+\mathbf{v})/2$. The displacement is the average velocity multiplied by the time interval:

$\mathbf{r}-\mathbf{r_0}=\left(\frac{\mathbf{v}+\mathbf{v_0}}{2} \right)t=\left(\frac{(\mathbf{v_0}+\mathbf{a}t)+\mathbf{v_0}}{2} \right)t$

           $=\mathbf{v_0}t+\frac{1}{2}\mathbf{a}t^2$

Or, $\mathbf{r}=\mathbf{r_0}+\mathbf{v_0}t+\frac{1}{2}\mathbf{a}t^2$  

It can be easily verified that the derivative of Eq. (4.34a), i.e. $\frac{d \mathbf{r}}{dt}$  gives Eq.(4.33a) and it also satisfies the condition that at t = 0, r = r$_0$. Equation (4.34a) can be written in component form as

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

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

One immediate interpretation of Eq.(4.34b) is that the motions in x- and y-directions can be treated independently of each other. That is, motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions. This is an important result and is useful in analysing motion of objects in two dimensions. A similar result holds for three dimensions. The choice of perpendicular directions is convenient in many physical situations, as we shall see in section 4.10 for projectile motion.

Example 1 

A particle starts from origin at t = 0 with a velocity 5.0$\mathbf{i}$ m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0$\mathbf{i}$ + 2.0$\mathbf{i}$) m/s$^2$. (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 m? (b) What is the speed of the particle at this time?

Answer

From Eq. (4.34a) for r$_0$ = 0, the position of the particle is given by

$\mathbf{r}(t)=\mathbf{v_0}t+\frac{1}{2}\mathbf{a}t^2$

$\mathbf{r}(t)=5.0\mathbf{i}+\frac{1}{2}(3.0\mathbf{i}+2.0 \mathbf{j})t^2$

         $=(5.0t+1.5t^2)\mathbf{i}+1.0t^2 \mathbf{j}$

Therefore, $x(t)=5.0t+1.5t^2$, $y=+1.0t^2$ 

Given x (t) = 84 m, t = ?

$5.0t+1.5t^2=84$ ⇒ t = 6 s

At t = 6 s, y = 1.0(6)$^2$ = 36.0 m

Now, the velocity $\mathbf{v}=\frac{d\mathbf{r}}{dt}=(5.0 +3.0t)\mathbf{i}+2.0t \mathbf{j}$

At t = 6 s, $\mathbf{v}=23.0 \mathbf{i}+12.0 \mathbf{j}$

speed $=|\mathbf{v}|=\sqrt{23^2+12^2} \cong$ 26 m.s$^{-1}$

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