Question

a hundred metre sprinter increases her speed from rest uniformly at the rate of 1mper sec square up-to three quarters of total run and covers the last quarter with uniform speed up. How much time does she take to cover the first half and the second half of the run?.

Answer 1

Answer:

Explanation:

It simpler by understand it diagram.

Answer 2

Answer:

Time to cover first half is 10seconds and Time to cover second half is 4.288seconds.

Explanation:

Given:

Total distance, S=100meter

First three quarter, S₁ =75m

Second quarter, $\mathrm{S}_{2}=25 \mathrm{~m}$

Acceleration, $a=1 \mathrm{~ms}^{-2}$

Initial velocity, $u=0$

Displacement ,S=

                        $\begin{aligned}&\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2} \\&\mathrm{t}_{1}=\sqrt{\frac{2 \mathrm{~S}_{1}}{\mathrm{a}}}=\sqrt{2 \times 75}=5 \sqrt{6} \mathrm{sec}\end{aligned}$

Velocity achieve in $3^{\text {rd }} quarter. v_{1}=u+ at =5 \sqrt{6} \mathrm{~ms}^{-1}$

In last one fourth quarter,

$\mathrm{t}_{2}=\frac{\mathrm{S}_{2}}{\mathrm{v}}=\frac{25}{5 \sqrt{6}}=\frac{5}{\sqrt{6}} \mathrm{sec}$

Total time T=

                        $\mathrm{T}=\mathrm{t}_{1}+\mathrm{t}_{2}=5 \sqrt{6}+\frac{5}{\sqrt{6}}=\frac{35}{\sqrt{6}} \mathrm{sec}$

(a) Time to cover first half

                        $\begin{aligned}&\mathrm{S}_{3}=\mathrm{ut}_{3}+\frac{1}{2} \mathrm{at}_{3}^{2} \\&\mathrm{t}_{3}=\sqrt{\frac{2 \mathrm{~S}_{3}}{\mathrm{a}}}=\sqrt{2 \times 50}=10 \mathrm{sec}\end{aligned}$

(b) Time to cover second half,

                        $\mathrm{t}_{4}=\mathrm{T}-\mathrm{t}_{3}=\frac{35}{\sqrt{6}}-10=4.288 \mathrm{sec}$