2.5 Constant acceleration (application)  (Page 2/2)

Problem : The coordinates of a particle (m/s) in a plane at a given time “t” is $2t,t^2$ . Find (i) path of motion (ii) velocity at time “t” and (iii) acceleration at time “t”.

Solution : Clearly, the position of the particle is a function of time and the particle moves in a two dimensional xy - plane. Here,

$$\begin{array}{l}x=2t\\ y=t^2\end{array}$$

In order to find the relation between “x” and “y”, we substitute “t” from the first equation in to second as :

$$\begin{array}{l}y={\left(\frac{x}{2}\right)}^2=\frac{x^2}{4}\\ \Rightarrow x^2=4y\end{array}$$

Hence, path of motion is parabolic. Now, the position vector is :

$$\begin{array}{l}\mathbf{r}=2t\mathbf{i}+t^2\mathbf{j}\end{array}$$

Differentiating with respect to time, the velocity of the particle is :

$$\begin{array}{l}\mathbf{v}=\frac{đ\mathbf{r}}{đt}=2\mathbf{i}+2t\mathbf{j}m/s\end{array}$$

Further differentiating with respect to time, the acceleration of the particle is :

$$\begin{array}{l}\mathbf{a}=\frac{đ\mathbf{v}}{đt}=2\mathbf{j}m/s^2\end{array}$$

Problem : A block is released from rest on a smooth inclined plane. If $S_n$ denotes the distance traveled by it from t = n - 1 second to t = n seconds, then find the ratio :

Motion along an incline

$$\begin{array}{l}A=\frac{S_n}{S_{n+1}}\end{array}$$

Solution : It must be noted that the description of linear motion is governed by the equations of motion whether particle moves on a horizontal surface (one dimensional description) or on an inclined surface (two dimensional description). Let us orient our coordinates so that the motion can be treated as one dimensional unidirectional motion. This allows us to use equations of motion in scalar form,

$$\begin{array}{l}\Rightarrow S_n=u+\frac{a}{2}(2n-1)\end{array}$$

Here, u = 0, thus

$$\begin{array}{l}\Rightarrow S_n=\frac{a}{2}(2n-1)\end{array}$$

Following the description of term Sn as given by the question, we can define Sn+1 as the linear distance from t = n second to t = n + 1 seconds. Thus, substituting “n” by “n+1” in the formulae, we have :

$$\begin{array}{l}\Rightarrow S_{n+1}=\frac{a}{2}(2n+2-1)\\ \Rightarrow S_{n+1}=\frac{a}{2}(2n+1)\end{array}$$

The required ratio is :

$$\begin{array}{l}\Rightarrow A=\frac{S_n}{S_{n+1}}=\frac{(2n-1)}{(2n+1)}\end{array}$$

Problem : A force of 2 N is applied on a particle of mass 1 kg, which is moving with a velocity 4 m/s in a perpendicular direction. If the force is applied all through the motion, then find displacement and velocity after 2 seconds.

Solution : It is a two dimensional motion, but having a constant acceleration. Notably, velocity and accelerations are not in the same direction. In order to find the displacement at the end of 2 seconds, we shall use algebraic method. Let the direction of initial velocity and acceleration be along “x” and “y” coordinates (they are perpendicular to each other). Also, let “A” be the initial position and “B” be the final position of the particle. The displacement between A (position at time t =0) and B (position at time t = 2 s) is given as :

$$\begin{array}{l}\mathbf{AB}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{t}^2\end{array}$$

For time t = 2 s,

$$\begin{array}{l}{\mathbf{AB}}_{\mathrm{t=2}}=2\mathbf{u}+\frac{1}{2}\mathbf{a}x{2}^2=2(\mathbf{u}+\mathbf{a})\end{array}$$

This is a vector equation involving sum of two vectors at right angles. According to question,

$$\begin{array}{l}u=4m/s;a=\frac{F}{m}=\frac{2}{1}=2m/s^2\end{array}$$

Since u and a perpendicular to each other, the magnitude of the vector sum ( u + a ) is :

$$\begin{array}{l}|\mathbf{u}+\mathbf{a}|=\surd (u^2+a^2)=\surd (4^2+2^2)=2\surd 5\end{array}$$

Hence, magnitude of displacement is :

$$\begin{array}{l}{\mathrm{AB}}_{\mathrm{t=2}}=2|\mathbf{u}+\mathbf{a}|==2x2\surd 5=4\surd 5m\end{array}$$

Let the displacement vector makes an angle “θ” with the direction of initial velocity.

$$\begin{array}{l}\tan \theta =\frac{a}{u}=\frac{2}{4}=\frac{1}{2}\end{array}$$

Let the direction of initial velocity and acceleration be along “x” and “y” coordinates (they are perpendicular to each other). Then,

$$\begin{array}{l}u=4\mathbf{i}\\ a=2\mathbf{j}\end{array}$$

Using equation of motion for constant acceleration, the final velocity is :

$$\begin{array}{l}\mathbf{v}=\mathbf{u}+\mathbf{a}t\\ \Rightarrow \mathbf{v}=4\mathbf{i}+2\mathbf{j}x1=4(\mathbf{i}+\mathbf{j})\end{array}$$

The magnitude of velocity is :

$$\begin{array}{l}v=\left|\mathbf{v}\right|=4\surd (1^2+1^2)=4\surd 2m\end{array}$$

Let the final velocity vector makes an angle “θ” with the direction of initial velocity.

$$\begin{array}{l}\tan \theta =\frac{1}{1}=1\\ \theta =45°\end{array}$$