Answer:
![3.8\text{m}/ \text{s}^2](https://img.qammunity.org/2020/formulas/physics/high-school/n7jd9wi8mwi09tgp25wv624wi6z0shq6yq.png)
Step-by-step explanation:
Given: At the starting gun, a runner accelerates at 1.9 m/s2 for 5.2 s. The runner’s acceleration is zero for the rest of the race.
To Find: the speed of the runner at t = 2.0 s.
Solution:
initial speed of runner(
) =
![\text{m}/ \text{s}^2](https://img.qammunity.org/2020/formulas/physics/high-school/4iy2bevzn3mnswknu8g628frl3ru0o3jcl.png)
acceleration for first 5.2s of race=
![\text{m}/ \text{s}^2](https://img.qammunity.org/2020/formulas/physics/high-school/4iy2bevzn3mnswknu8g628frl3ru0o3jcl.png)
to find speed of runner at ,
![\text{t}=2\text{s}](https://img.qammunity.org/2020/formulas/physics/high-school/vk2l3vnu8a2k8mykxioo8p6ee3vpsxhtjv.png)
Using first equation of motion,
![\text{v}=\text{u}+\text{at}](https://img.qammunity.org/2020/formulas/physics/high-school/rkjpzcx3y05e2jq4we1e50zh0o34iblcl4.png)
putting values in the equation
![\text{v}=0+1.9*2](https://img.qammunity.org/2020/formulas/physics/high-school/dmyaoum2pemqrg03ql1sbytpj9uxxa8sv0.png)
![\text{m}/ \text{s}^2](https://img.qammunity.org/2020/formulas/physics/high-school/4iy2bevzn3mnswknu8g628frl3ru0o3jcl.png)
So,
The speed of the runner at
, is
![3.8\text{m}/ \text{s}^2](https://img.qammunity.org/2020/formulas/physics/high-school/n7jd9wi8mwi09tgp25wv624wi6z0shq6yq.png)