Question

A man is driving his car with constant acceleration on a highway. He crosses two check posts with speed $30\sqrt{3}m/s$ and $10\sqrt{3}m/s$. The speed of the car when he is exactly $1/{{3}^{rd}}$ of total distance from pole $\text{I}$ is
$\begin{align}
  & a)35\sqrt{2}m/s \\
 & b)50m/s \\
 & c)50\sqrt{2}m/s \\
 & d)10\sqrt{35}m/s \\
\end{align}$

Answer 1

Hint: In the above question we are basically asked to determine the speed of the car when it reaches one third of the total distance from pole one. Let us say the car crosses the second post with speed of $10\sqrt{3}m/s$ and the first post with $30\sqrt{3}m/s$. It is also given to us that the car is under a uniform acceleration. Therefore we will use Newton’s third kinematic equation to determine the speed of the car when it is at a distance $1/{{3}^{rd}}$ of the total distance between the two posts from the post one.

Formula used:
${{V}^{2}}-{{U}^{2}}=2aS$

Complete step by step answer:
Let us say a body moves with uniform acceleration ‘a’. If the initial speed of the body is ‘U’, than the speed of the body ‘V’ when it covers a distance S from Newton’s third kinematic equation is given by,
${{V}^{2}}-{{U}^{2}}=2aS$
It is given to us in the question that the car moves from post two to one with uniform acceleration a. The car crosses the second post with speed of $10\sqrt{3}m/s$ and the first post with $30\sqrt{3}m/s$. Let us say the distance between the two is S. After substituting the velocities of the car when it passes across the two check posts, from Newton’s third kinematic equation we get,
$\begin{align}
  & {{V}^{2}}-{{U}^{2}}=2aS \\
 & \Rightarrow {{\left( 30\sqrt{3} \right)}^{2}}-{{\left( 10\sqrt{3} \right)}^{2}}=2aS...(1) \\
\end{align}$
Further we are asked to determine the speed(V) of the car when it is at a distance of one third of S, from pole one. Hence using the above kinematic equation we get,
$\begin{align}
  & {{V}^{2}}-{{U}^{2}}=2aS \\
 & \Rightarrow {{\left( V \right)}^{2}}-{{\left( 30\sqrt{3} \right)}^{2}}=2a\dfrac{S}{3}...(2) \\
\end{align}$
Dividing equation 1 by 2 we get,
$\begin{align}
  & \dfrac{{{\left( 30\sqrt{3} \right)}^{2}}-{{\left( 10\sqrt{3} \right)}^{2}}}{{{\left( V \right)}^{2}}-{{\left( 30\sqrt{3} \right)}^{2}}}=\dfrac{2aS}{2a\dfrac{S}{3}} \\
 & \Rightarrow \dfrac{2700-300}{{{\left( V \right)}^{2}}-2700}=3 \\
 & \Rightarrow 3\left[ {{\left( V \right)}^{2}}-2700 \right]=2400 \\
 & \Rightarrow 3{{\left( V \right)}^{2}}=2400+8100=10500 \\
 & \Rightarrow {{\left( V \right)}^{2}}=\dfrac{10500}{3}=3500 \\
 & \therefore V=10\sqrt{35}m/s \\
\end{align}$

Hence the correct answer of the above question is option D.

Note:
In the question it is clearly mentioned that the car is accelerating uniformly. Therefore we can use Newton’s kinematic equations to determine its motion parameters. It is also to be noted that the term pole mentioned in the question represents the corresponding check post.