question_answer

A point moving in a straight line travels in its second, fifth and eleventh seconds of motion 16 m, 28 m and 52 m respectively. Prove that the point is moving with constant acceleration. Also find the total distance moved in 10 seconds.

Answer:

                Let the initial velocity be u and the acceleration be a. Distance travelled in the nth second by a particle moving in a straight line with a constant acceleration is \[{{\text{D}}_{\text{n}}}=\text{u}+\frac{a}{2}\left( \text{2n}\text{1} \right)\] So, \[\text{16 }=\text{ u}+\frac{a}{2}~\left( \text{2}\times \text{2}\text{1} \right)\]                                                                        .....(i) \[\text{28}=\text{u}+~\frac{a}{2}\left( \text{2}\times \text{5}\text{1} \right)\]                                                                                 ......(ii) and \[\text{52}=+\text{a}/\text{2 }\left( \text{2}\times \text{11}\text{1} \right)\]                                                                           ......(iii) On solving (i), (ii) and (iii), we get, \[\text{a}=\text{4m}{{\text{s}}^{-\text{2}}};\text{ u}=\text{1}0\text{m}{{\text{s}}^{-\text{1}}}\] Now, \[\text{s}=\text{ut}+\text{1}/\text{2 a}{{\text{t}}^{\text{2}}}=\text{1}0\times \text{1}0+\text{1}/\text{2}\times \text{4}\times \text{1}{{0}^{\text{2}}}=\text{3}00\text{m}\].