question_answer

A 100 m sprinter increases her speed from rest uniformly at the rate of \[\text{1}.\text{5 m}{{\text{s}}^{-\text{2}}}\] up to three quarters of the total run and covers the last quarter with uniform speed. How much time does she take to cover the first half and the second half of the run?

Answer:

                In Fig. 2(HT).22, total distance AD = 100 m. Three quarter distance = AC = 75 m and half of the total distance = AB = 50 m. Therefore distance Taking motion of sprinter from A to B, we have ; u = 0; \[\text{a }=\text{1}.\text{5 m}{{\text{s}}^{-\text{2}}}\]; t = ?;.s = 50 m As \[~\text{s}=\text{ut}+\frac{1}{2}\text{a}{{\text{t}}^{\text{2}}}\]; \[\text{5}0=0\times \text{t }+\frac{1}{2}\text{1}.\text{5 }\times {{\text{t}}^{\text{2}}}\] or            \[\text{t }={{\left( \text{2}\times \text{5}0/\text{1}.\text{5} \right)}^{\text{1}/\text{2}}}=\text{8}.\text{165s}\] Taking motion of sprinter from A to C; u = 0; \[\text{a }=\text{ 1}.\text{5 m}{{\text{s}}^{-\text{2}}}\]; t = ? s = 75 m; v = ? As           \[\text{s}=\text{ut}+\frac{1}{2}\text{a}{{\text{t}}^{\text{2}}}\]; \[\text{75}=0\times \text{ t }+\frac{1}{2}\times \text{ 1}.\text{5 }\] or            \[\text{t}={{\left( \text{75}\times \text{2}/\text{1}.\text{5} \right)}^{\text{1}/\text{2}}}=\text{1}0\text{s}\]. Also       \[\text{v}=\text{u}+\text{at}=0+\text{1}.\text{5}\times \text{1}0=\text{15 m}{{\text{s}}^{-\text{1}}}\]. Therefore, time taken from B to C = 10 - 8.165 = 1.835 s Taking motion of sprinter from C to D; \[\text{u }=\text{ 15 m}{{\text{s}}^{-\text{1}}}\]; t = ? s = 25 m; a = 0 As,          \[\text{s }=\text{ ut }+\frac{1}{2}\text{a}{{\text{t}}^{\text{2}}}\]; \[\text{25}=\text{15}\times \text{t}+\frac{1}{2}\left( 0 \right)\times {{\text{t}}^{\text{2}}}=\text{15t}\] or            t = 25/15 = 1.667 s time from B to D = 1.835 + 1.667 = 3.502 s.