AP EAMCET · PHYSICS · Motion In One Dimension
A body starts from rest and moves with uniform acceleration. If the distance travelled by it in the first \(2 \mathrm{~s}\) is \(x_1\) and in the next \(2 \mathrm{~s}\) is \(x_2\), then \(x_1\) and \(x_2\) are related as
- A \(x_1=x_2\)
- B \(x_1=2 x_2\)
- C \(2 x_1=x_2\)
- D \(3 x_1=x_2\)
Answer & Solution
Correct Answer
(D) \(3 x_1=x_2\)
Step-by-step Solution
Detailed explanation
If \(a\) be the acceleration of the body, then distance travelled by the body in \(2 \mathrm{~s}\) is given as, \[ \begin{gathered} x_1=u t+\frac{1}{2} a t^2=0 \times 2+\frac{1}{2} a \times(2)^2 \\ x_1=2 a \end{gathered} \] Velocity of body at the end of \(2 \mathrm{~s}\) is…
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