question_answer

An electron starting from rest has a velocity that increases linearly with time that is\[\text{v}=\text{kt}\], where\[\text{k}=\text{2 m}/{{\text{s}}^{\text{2}}}\]. What will be the distance covered in first 3 seconds of its motion?

Answer:

                Here, \[s=0,\text{v }=\text{1}00\text{ m}{{\text{s}}^{-\text{1}}};\text{a }=\text{ }-\text{1}0\text{ m}{{\text{s}}^{-\text{2}}},\text{t }=?\] Using the relation, \[\text{s}=\text{ut }+\frac{1}{2}\text{a}{{\text{t}}^{\text{2}}}\], we have \[0=\text{1}00\text{t}+~\frac{1}{2}\left( -\text{1}0 \right){{\text{t}}^{\text{2}}}\]or \[\text{t}=\text{2s}\].