Search any question & find its solution
Question:
Answered & Verified by Expert
A body starts from rest and moves with constant acceleration for \( t \) s. It travels a distance \( x_{1} \) in
first half of time and \( x_{2} \) in next half of time, then
Options:
first half of time and \( x_{2} \) in next half of time, then
Solution:
1962 Upvotes
Verified Answer
The correct answer is:
\( x_{2}=3 x_{1} \)
Initial velocity $=0$ and acceleration is constant.
Distance travelled in first half of time $=\chi_{1}$
Distance travelled in next half of time $=x_{2}$
Therefore, acceleration $=\frac{x_{2}-x_{1}}{t^{2}} \rightarrow$ (1)
Now, $x_{1}=\frac{1}{2}$ at $^{2}$ because initial velocity is zero. Substitute in Eq. (1), we get
$$
\begin{array}{l}
a=\frac{x_{2}-\frac{1}{2} a t^{2}}{t^{2}} \\
\Rightarrow a t^{2}=x_{2}-\frac{1}{2} a t^{2} \\
\Rightarrow x_{2}=a t^{2}+\frac{1}{2} a t^{2} \\
\Rightarrow x_{2}=3 \times \frac{1}{2} a t^{2}
\end{array}
$$
Thus $x_{2}=3 x_{1}$
Distance travelled in first half of time $=\chi_{1}$
Distance travelled in next half of time $=x_{2}$
Therefore, acceleration $=\frac{x_{2}-x_{1}}{t^{2}} \rightarrow$ (1)
Now, $x_{1}=\frac{1}{2}$ at $^{2}$ because initial velocity is zero. Substitute in Eq. (1), we get
$$
\begin{array}{l}
a=\frac{x_{2}-\frac{1}{2} a t^{2}}{t^{2}} \\
\Rightarrow a t^{2}=x_{2}-\frac{1}{2} a t^{2} \\
\Rightarrow x_{2}=a t^{2}+\frac{1}{2} a t^{2} \\
\Rightarrow x_{2}=3 \times \frac{1}{2} a t^{2}
\end{array}
$$
Thus $x_{2}=3 x_{1}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.