KCET · Physics · Gravitation
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
- A \( x_{2}=x_{1} \)
- B \( x_{2}=2 x_{1} \)
- C \( x_{2}=3 x_{1} \)
- D \( x_{2}=4 x_{1} \)
Answer & Solution
Correct Answer
(C) \( x_{2}=3 x_{1} \)
Step-by-step Solution
Detailed explanation
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
\(a=\frac{x_{2}-\frac{1}{2} a t^{2}}{t^{2}} \)
\(\Rightarrow a t^{2}=x_{2}-\frac{1}{2} a t^{2} \)
\(\Rightarrow x_{2}=a t^{2}+\frac{1}{2} a t^{2} \)
\(\Rightarrow x_{2}=3 \times \frac{1}{2} a t^{2}\)
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
\(a=\frac{x_{2}-\frac{1}{2} a t^{2}}{t^{2}} \)
\(\Rightarrow a t^{2}=x_{2}-\frac{1}{2} a t^{2} \)
\(\Rightarrow x_{2}=a t^{2}+\frac{1}{2} a t^{2} \)
\(\Rightarrow x_{2}=3 \times \frac{1}{2} a t^{2}\)
Thus \(x_{2}=3 x_{1}\)
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