Answer:
The value is
![s = 306.6 \ m](https://img.qammunity.org/2021/formulas/physics/high-school/o99pe9ymwsgdf5h7g36lhrnfrmbuvypry4.png)
Step-by-step explanation:
From the question we are told that
The acceleration is
![a = 2.52 \ m/s](https://img.qammunity.org/2021/formulas/physics/high-school/ai9np5qz2p7zn3buqlb9ghw0m0f3pi718b.png)
The speed to attain is
![v = 39.2 \ m/s](https://img.qammunity.org/2021/formulas/physics/high-school/5hcok9tosdyu0tbgkf4d1hct6w092iozhl.png)
Generally the time taken is mathematically represented as
![t = (v)/(a)](https://img.qammunity.org/2021/formulas/physics/high-school/ug7t40vsn6kt83keh6bo87fvcaza6w3nmw.png)
=>
![t = (39.2)/(2.52)](https://img.qammunity.org/2021/formulas/physics/high-school/pndd54zhcwft9118cu14ruvpa9p1bajqlc.png)
=>
![t= 15.6 \ s](https://img.qammunity.org/2021/formulas/physics/high-school/d1cj4zm8nn22j4xmv0pyqc4b6qbyxwk338.png)
Generally from the kinematic equation we have that
![s = ut + (1)/(2)at^2](https://img.qammunity.org/2021/formulas/physics/high-school/xrbe6fp2nwpze8ak7i4g2sdc2cokv3ewgy.png)
Here the initial velocity is zero i.e u = 0 \ m/s (it started from rest)
So
![s = 0* 15.6 + (1)/(2) * 2.52 * (15.6)^2](https://img.qammunity.org/2021/formulas/physics/high-school/4dgzktv494tr2i4iwltig0gvlmlo58k75h.png)
=>
![s = 306.6 \ m](https://img.qammunity.org/2021/formulas/physics/high-school/o99pe9ymwsgdf5h7g36lhrnfrmbuvypry4.png)