An athlete runs a distance of $1500 m$ in the following manner. $(i)$ Starting from rest, he accelerates himself uniformly at $2 m s^{- 2}$ till he covers a distance of $900 m$ . $(i i)$ He, the run remaining distance of $600 c m$ at the uniform speed developed. Calculate the time taken by the athlete to cover the two parts of the distance covered. Also find the time, when he is at the centre of the track.

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Solution

The situation is shown in Fig.
Fig.
For motion between times $t = 0$ and $t_{2}$ :
$u = 0, a = 2 m s^{- 2}, s = 900 m, v = v_{2}$ (say)
As $v^{2} - u^{2} = 2 a s$
$∴ v_{2}^{2} - 0^{2} = 2 \times 2 \times 900$ or $v_{2} = 60 m s^{- 1}$
Time, $t_{2} = \frac{v - u}{a} = \frac{60 - 0}{2} = 30 s$ .
For motion between times $t_{2}$ and $t_{3}$ : This motion ???? at a constant speed of $60 m s^{- 1}$ .
$∴$ Time taken, $t_{3} - t_{2} = \frac{1500 - 900}{60} = 10 s$
For motion between time $t = 0$ and $t_{1}$ :
$u = 0, s = 750 m, a = 2 m s^{- 2}, t = t_{1}$ (say)
As $s = u t + \frac{1}{2} a t^{2}$ $∴ 750 = 0 \times t_{1} + \frac{1}{2} \times 2 \times t_{1}^{2}$
$t_{1} = \sqrt{750} = 27.4 s$ .

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