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.
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