TS EAMCET · Physics · Motion In One Dimension
A body moving with uniform acceleration, travels a distance of 25 m in the fourth second and 37 m in the sixth second. The distance covered by the body in the next two seconds is
- A 63 m
- B 84 m
- C 49 m
- D 92 m
Answer & Solution
Correct Answer
(D) 92 m
Step-by-step Solution
Detailed explanation
For uniformly accelerated motion, \(S_n=u+(2 n-1) \frac{a}{2}\) \(\therefore \quad 25=u+(2 \times 4-1) \frac{a}{2}=u+\frac{7}{2} a\) ...(i) Again, \(37=u+(2 \times 6-1) \frac{a}{2}=u+\frac{11}{2} a\) ...(ii) From eq (i) and (ii), we get…
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