Question

Two particles are travelling due east with different velocities. However, after $4s$ they have the same velocities. During this $4s$ interval, average acceleration of $1^{st}$ particle is $2m{s}^{-2}$ due east while that of $2^{nd}$ particle is $4m{s}^{-2}$ due east. Determine the difference in their speeds at the beginning of $4s$ duration and also fine which one is moving faster initially.

Answer 1

Step 1: Given Data

The average acceleration of $1^{st}$ particle, $a_1=2m{s}^{-2}$.

The average acceleration of $2^{nd}$ particle, $a_2=4m{s}^{-2}$.

The time duration after which both particles have the same velocities, $t=4s$.

Step 2: Determine which particle was faster than the other

Since the two particles have the same velocities after $t=4s$ ,

Let us assume,

1. the final velocity attained by $1^{st}$ particle $=v_{1}m{s}^{-1}$
2. the final velocity attained by $2^{nd}$ particle $=v_{2}m{s}^{-1}$
3. the initial velocity of $1^{st}$ particle $=u_{1}m{s}^{-1}$
4. the initial velocity of $2^{nd}$ particle $=u_{2}m{s}^{-1}$

From the given data we can clearly see, $a_2>a_1$. Thus, $1^{st}$ particle was moving faster and $2^{nd}$ particle that is why their final velocities became equal at $4s$

Step 3: Determine the final velocities of the particles

After $4s$ both the particles have the same velocities, Therefore, $v_2=v_1$

As we know that the first equation of motion is $v=u+at$. On substituting the given data in the above equation we get,

$$\begin{gathered} \Rightarrow u_1+2\times 4=u_2+4\times 4 \\ \Rightarrow u_1+8=u_2+16 \\ \Rightarrow u_1-u_2=16-8 \\ \Rightarrow u_1-u_2=8m{s}^{-1} \end{gathered}$$

Final Answer:

The difference between the speed of the two particles is $\mathbf{8}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$ at the beginning of $\mathbf{4}\mathbf{}\mathit{s}$ and ${\mathbf{1}}^{\mathbf{s}\mathbf{t}}$ particle with smaller acceleration was moving faster.