if 100m sprinters accelerate from rest for 3.5s at 2.8m/s^2, how far have they run to this

For the first part use s = ut + (1/2)at², where s = distance, u = initial speed, a = acceleration, t = time:

$${\mathtt{s}} = {\mathtt{0}}{\mathtt{\,\times\,}}{\mathtt{3.5}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{2.8}}{\mathtt{\,\times\,}}{{\mathtt{3.5}}}^{{\mathtt{2}}} \Rightarrow {\mathtt{s}} = {\mathtt{17.15}}$$

So distance = 17.15m

 

For the second part find the speed from v = u + at:

$${\mathtt{v}} = {\mathtt{0}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.8}}{\mathtt{\,\times\,}}{\mathtt{3.5}} \Rightarrow {\mathtt{v}} = {\mathtt{9.8}}$$

v = 9.8m/s

 

At this speed the sprinter will cover the remaining distance in time t2 = (100 - 17.15)/9.8 seconds

$${\mathtt{t2}} = {\frac{\left({\mathtt{100}}{\mathtt{\,-\,}}{\mathtt{17.15}}\right)}{{\mathtt{9.8}}}} \Rightarrow {\mathtt{t2}} = {\mathtt{8.454\: \!081\: \!632\: \!653\: \!061\: \!2}}$$

t2 ≈ 8.45 s

 

So, total time for the 100m is 3.5 + 8.45 seconds = 11.95 seconds

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