Motion in plane with uniform (Constant) acceleration

An object is said to be moving with uniform acceleration, if its velocity vector undergoes the same change in the same interval of time (however small).
Let an object is moving in \(X Y\)-plane and its acceleration a is constant. At time \(t=0\), the velocity of an object be \(\mathbf{v}_0\) (say) and \(\mathbf{v}\) be the velocity at time \(t\).
According to definition of average acceleration, we get
\(
\mathbf{a}=\frac{\mathbf{v}-\mathbf{v}_0}{t-0}=\frac{\mathbf{v}-\mathbf{v}_0}{t} \Rightarrow \mathbf{v}=\mathbf{v}_0+\mathbf{a} t
\)
In terms of rectangular components, we can express it as
\(
v_x=v_{0 x}+a_x t \text { and } v_y=v_{0 y}+a_y t
\)
It can be concluded that, each rectangular component of velocity of an object moving with uniform acceleration in a plane depends upon time as if it were the velocity vector of one dimensional uniformly accelerated motion.

Path of particle under constant acceleration

Now, we can also find the position vector \((\mathbf{r})\). Let \(\mathbf{r}_0\) and \(\mathbf{r}\) be the position vectors of the particle at time \(t=0\) and \(t=t\) and their velocities at these instants be \(\mathbf{v}_0\) and \(\mathbf{v}\), respectively. Then, the average velocity is given by
\(
\mathbf{v}_{\mathrm{av}}=\frac{\mathbf{v}_0+\mathbf{v}}{2}
\)
Displacement is the product of average velocity and time interval. It is expressed as
\(
\mathbf{r}-\mathbf{r}_0=\left(\frac{\mathbf{v}+\mathbf{v}_0}{2}\right) t=\left[\frac{\left(\mathbf{v}_0+\mathbf{a} t\right)+\mathbf{v}_0}{2}\right] t
\)
\(
\mathbf{r}-\mathbf{r}_0=\mathbf{v}_0 t+\frac{1}{2} \mathbf{a} t^2
\)
\(
\mathbf{r}=\mathbf{r}_0+\mathbf{v}_0 t+\frac{1}{2} \mathbf{a} t^2
\)
In terms of rectangular components, we get
\(
x \hat{\mathbf{i}}+y \hat{\mathbf{j}}=x_0 \hat{\mathbf{i}}+y_0 \hat{\mathbf{j}}+\left(v_{0 x} \hat{\mathbf{i}}+v_{0 y} \hat{\mathbf{j}}\right) t+\frac{1}{2}\left(a_x \hat{\mathbf{i}}+a_y \hat{\mathbf{j}}\right) t^2
\)
Now, equating the coefficients of \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\),
\(
x=x_0+v_{0 x} t+\frac{1}{2} a_x t^2 \ldots . . . \text { along } X \text {-axis }
\)
\(
\text { and } \quad y=y_0+v_{0 y} t+\frac{1}{2} a_y t^2 \ldots . . . \text { along } Y \text {-axis }
\)
Note: Motion in a plane (two dimensional motion) can be treated as two separate simultaneous one dimensional motions with constant acceleration along two perpendicular directions.

Example 22: (i) What does \(\left|\frac{d \mathbf{v}}{d t}\right|\) and \(\frac{d|\mathbf{v}|}{d t}\) represent?
(ii) Can these be equal?
(iii) Can \(\frac{d|\mathbf{v}|}{d t}=0\) while \(\left|\frac{d \mathbf{v}}{d t}\right| \neq 0\) ?
(iv) \(\operatorname{Can} \frac{d|\mathbf{v}|}{d t} \neq 0\) while \(\left|\frac{d \mathbf{v}}{d t}\right|=0\) ?

Solution: (i) \(\left|\frac{d \mathbf{v}}{d t}\right|\) is the magnitude of total acceleration. While \(\frac{d|\mathbf{v}|}{d t}\) represents the time rate of change of speed (called the tangential acceleration, a component of total acceleration) as \(|\mathbf{v}|=v\).
(ii) These two are equal in case of one dimensional motion (without change in direction).
(iii) In case of uniform circular motion, speed remains constant while velocity changes.
Hence, \(\frac{d|\mathbf{v}|}{d t}=0\) while \(\left|\frac{d \mathbf{v}}{d t}\right| \neq 0\).
(iv) \(\frac{d|\mathbf{v}|}{d t} \neq 0\) implies that speed of particle is not constant. Velocity cannot remain constant, if speed is changing. Hence, \(\left|\frac{d \mathbf{v}}{d t}\right|\) cannot be zero in this case. So, it is not possible to have \(\left|\frac{d \mathbf{v}}{d t}\right|=0\) while \(\frac{d|\mathbf{v}|}{d t} \neq 0\).

Example 23: \(A\) particle starts from origin at \(t=0\) with \(a\) velocity of \(15 \hat{\mathbf{i}} \mathrm{ms}^{-1}\) and moves in \(X Y\)-plane under the action of a force which produces a constant acceleration of \((15 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}) \mathrm{ms}^{-2}\). Find the \(y\)-coordinate of the particle at the instant when its \(x\)-coordinate is \(180 \mathrm{~m}\).

Solution: The position of the particle is given by
\(
\mathbf{r}(t)=\mathbf{v}_0 t+\frac{1}{2} \mathbf{a} t^2=15 \hat{\mathbf{i}} t+\frac{1}{2}(15 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}) t^2
\)
\(
\begin{aligned}
& =\left(15 t+7.5 t^2\right) \hat{\mathbf{i}}+10 t^2 \hat{\mathbf{j}} \\
\therefore \quad x(t) & =15 t+7.5 t^2 \text { and } y(t)=10 t^2
\end{aligned}
\)
If \(x(t)=180 \mathrm{~m}, t=\) ?
\(
180=15 t+7.5 t^2 \Rightarrow t=4 \mathrm{~s}
\)
\(\therefore y\)-coordinate, \(y(t)=10 \times 16=160 \mathrm{~m}\)

Example 24: An object has a velocity \(\mathbf{v}=(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{ms}^{-1}\) at time \(t=0 \mathrm{~s}\). It undergoes a constant acceleration \(a=(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}) m s^{-2}\) for \(4 s\). Then,
(i) find the coordinates of the object, if it is at origin at \(t=0\).
(ii) find the magnitude of its velocity at the end of \(4 \mathrm{~s}\).

Solution: (i) Here, initial position of the object,
\(
\begin{array}{l}
\qquad \mathbf{r}_0=x_0 \hat{\mathbf{i}}+y_0 \hat{\mathbf{j}}=0 \hat{\mathbf{i}}+0 \hat{\mathbf{j}} \\
\text { Initial velocity, } \mathbf{v}_0=v_{0_x} \hat{\mathbf{i}}+v_{0_y} \hat{\mathbf{j}}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}} \\
\text { Acceleration, } \quad \mathbf{a}=a_x \hat{\mathbf{i}}+a_y \hat{\mathbf{j}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}} \\
\text { and } t=4 \mathrm{~s}
\end{array}
\)
Let the final coordinates of the object be \((x, y)\). Then, according to the equation for the path of particle under constant acceleration,
\(
\begin{aligned}
x & =x_0+v_{0 x} t+\frac{1}{2} a_x t^2=0+2 \times 4+\frac{1}{2}(1) \times 4^2 \\
\Rightarrow \quad x & =16 \mathrm{~m} \text { and } y=y_0+v_{0 y} t+\frac{1}{2} a_y t^2 \\
& =0+4 \times 4+\frac{1}{2}(-3) \times 4^2 \Rightarrow y=-8 \mathrm{~m}
\end{aligned}
\)
Therefore, the object lies at \((16 \hat{\mathbf{i}}-8 \hat{\mathbf{j}})\) at \(t=4 \mathrm{~s}\).
(ii) Using equation
\(
\begin{array}{l}
\mathbf{v}=\mathbf{v}_0+\mathbf{a} t \\
\Rightarrow \quad \mathbf{v}=(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})+(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \times 4 \\
=(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})+(4 \hat{\mathbf{i}}-12 \hat{\mathbf{j}})=(2+4) \hat{\mathbf{i}}+(4-12) \hat{\mathbf{j}} \\
\Rightarrow \quad \mathbf{v}=6 \hat{\mathbf{i}}-8 \hat{\mathbf{j}} \\
\end{array}
\)
\(\therefore\) Magnitude of velocity, \(|\mathbf{v}|=\sqrt{6^2+8^2}=10 \mathrm{~ms}^{-1}\)
Its direction with \(X\)-axis, \(\theta=\tan ^{-1}\left(\frac{-8}{6}\right) \approx-53^{\circ}\)