AP EAMCET · PHYSICS · Motion In One Dimension
A body starting from rest at \(t=0\) moves along a straight line with a constant acceleration. At \(t=2 \mathrm{~s}\), the body reverses its direction keeping the acceleration same. The body returns to the initial position at \(t=t_0\), then \(t_0\) is
- A 4 s
- B \((4+2 \sqrt{2}) s\)
- C \((2+2 \sqrt{2}) s\)
- D \((4+4 \sqrt{2}) s\)
Answer & Solution
Correct Answer
(B) \((4+2 \sqrt{2}) s\)
Step-by-step Solution
Detailed explanation
According to the question, From first equation of the motion, \( v_1=u+a t_1 \Rightarrow v_1=2 a \) Firstly, body decelerate with acceleration to the point \(C\) and then reverse it's direction and accelerate with acceleration \(a\) to the point \(A\). Therefore for distance…
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