Question

A car accelerates from rest at a constant rate for first $10\text{ }s$ and covers a distance \[x\] . It converts a distance $y$ in next $10\text{ }s$ at the same acceleration. Which of the following is true?
(A) \[x=3y\]
(B) $y=3x$
(C) \[x=y\]
(D) $y=2x$

Answer 1

Hint: In the first case, take initial velocity zero because the car is at rest. Then find the final velocity using the given time, find the distance travelled in the first case.
In the second case, consider initial velocity(that is final velocity in first case),then find the distance travelled in second case. Compare the results from both the cases, we get a required relation.

Formula used
$\begin{align}
  & v=u+at \\
 & v\text{ is final velocity} \\
 & u\text{ is initial velocity} \\
 & a\text{ is acceleration} \\
 & t\text{ is time taken} \\
\end{align}$
$\text{Distance travelled is given by }5=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete step by step solution

First case: Initially, car at rest, it means.
          $\begin{align}
  & u\left( \text{initial velocity} \right)=0 \\
 & \text{Final velocity is given by} \\
 & \text{ }v=u+at \\
 & \text{ }v=0+at \\
 & \text{ }t=10s\text{ }a\text{ constant acceleration} \\
\end{align}$
          $\begin{align}
  & \text{ }V=10a \\
 & \left( x \right)\text{ distance travelled is given by,} \\
 & \text{ }s=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & \text{ }s=0+\dfrac{1}{2}a{{t}^{2}}\text{ } \\
\end{align}$
               $\text{ }x=50a$…………. (1)

In second case:
\[\begin{align}
  & \text{Initial velocity}=10a\text{ }\left( \text{ This is final velocity in first case} \right) \\
 & \text{ }v=u+at \\
 & \text{ }=100+10a \\
\end{align}\]
               \[\text{ }v=20a\] final velocity in second case
\[\begin{align}
  & \left( y \right)\text{ Distance travelled is given by} \\
 & s=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & \text{ }=\left( 10a \right)\left( 10 \right)+\dfrac{1}{2}a{{\left( 10 \right)}^{2}} \\
 & \text{ }=100a+50a \\
\end{align}\]
\[s=150a\]
\[y=150a\]…….. (2)
Divide equation (1) by (2)
$\begin{align}
  & \dfrac{x}{y}=\dfrac{50}{150} \\
 & \dfrac{x}{y}=\dfrac{1}{3} \\
 & y=3x\text{ option(b)} \\
 & \text{This is the required result}\text{.} \\
\end{align}$

Note: Discuss and derive the formula for
\[\text{Velocity time relation, }v=u+at\]
$\text{Position time relation, }s=ut+\dfrac{1}{2}a{{t}^{2}},\text{ (}s\text{ is distance covered)}$
$\text{Position velocity relation, }{{v}^{2}}-{{u}^{2}}=2\text{ }as$