Đáp án:

\(\begin{align}
  & {{a}_{1}}=0km/{{h}^{2}} \\ 
 & {{a}_{2}}=1296km/{{h}^{2}} \\ 
 & {{a}_{3}}=-5184km/{{h}^{2}} \\ 
\end{align}\)

Giải thích các bước giải:

\({{v}_{1}}=30km/h;{{t}_{1}}=30p=0,5h;{{t}_{2}}=100s;{{v}_{2}}=72km/h;S=500m\)

a> Gia tốc:
\(\begin{align}
  & {{a}_{1}}=0m/s \\ 
 & {{v}_{2}}={{v}_{1}}+{{a}_{2}}.{{t}_{2}}\Rightarrow {{a}_{2}}=\frac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}}=\frac{72-36}{100: 3600}=1296km/{{h}^{2}} \\ 
 & v_{3}^{2}-v_{2}^{2}=2{{a}_{3}}.S\Rightarrow {{a}_{3}}=\frac{v_{3}^{2}-v_{2}^{2}}{2S}=\frac{-{{72}^{2}}}{2.0,5}=-5184km/{{h}^{2}} \\ 
\end{align}\)

b> Thời gian đi từ C đên D
\({{v}_{3}}={{v}_{2}}+{{a}_{3}}.{{t}_{3}}\Rightarrow {{t}_{3}}=\dfrac{-72}{-5184}=\dfrac{1}{72}h\)

Tổng thời gian A-D: 

\(t={{t}_{1}}+{{t}_{2}}+{{t}_{3}}=0,5+\dfrac{1}{36}+\dfrac{1}{72}=\dfrac{13}{24}h=1950s\)

c> Quãng đường:
\(S={{v}_{1}}.{{t}_{1}}+{{v}_{1}}.{{t}_{2}}+\dfrac{1}{2}.a{}_{2}.t_{2}^{2}+0,5=36.0,5+36.\dfrac{1}{36}+\dfrac{1}{2}.1296.{{(\dfrac{1}{36})}^{2}}+0,5=20km\)

d> Quãng đường đi trong 1945s đầu:
\(\begin{align}
  & S'={{v}_{1}}.{{t}_{1}}+{{v}_{1}}.{{t}_{2}}+\frac{1}{2}.a{}_{2}.t_{2}^{2}+{{v}_{2}}.({{t}_{3}}-\dfrac{1}{720})+\frac{1}{2}.{{a}_{3}}.{{({{t}_{3}}-\dfrac{1}{720})}^{2}} \\ 
 & =36.0,5+36.\dfrac{1}{36}+\dfrac{1}{2}.1296.{{(\frac{1}{36})}^{2}}+72.(\dfrac{1}{72}-\dfrac{1}{720})-\dfrac{1}{2}.5184.{{(\frac{1}{72}-\dfrac{1}{720})}^{2}}=19,995km \\ 
\end{align}\)

Quãng đường đi trong 5s cuối: 

\({{S}_{5}}=20-19,995=0,005km=5m\)