\(v_0=72\)km/h=20m/s

Gia tốc:  \(v^2-v^2_0=2aS\)

\(\Rightarrow a=\dfrac{v^2-v^2_0}{2S}=\dfrac{0-20^2}{2\cdot40}=-5\)m/s²

Hệ số ma sát:

\(a=-\mu\cdot g\Rightarrow\mu=-\dfrac{a}{g}=-\dfrac{-5}{10}=0,5\)

a. Ta có: \(\left\{{}\begin{matrix}s=v_0t+\dfrac{1}{2}at^2\Leftrightarrow24=v_0\cdot4+\dfrac{1}{2}\cdot a\cdot4^2\Leftrightarrow24=v_0\cdot4+8a_{\left(1\right)}\\v_0+at=0\Leftrightarrow v_0=-a\cdot4_{\left(2\right)}\end{matrix}\right.\)

Thay (2) vào (1): \(24=8a-16a\)

\(\Rightarrow a=-3\left(\dfrac{m}{s^2}\right)\)

\(\Rightarrow v_0=-at=-3\cdot-4=12\left(\dfrac{m}{s}\right)\)

b. \(F_{ham}=ma=2000\cdot-3=-6000\left(N\right)\)

c. \(F'_{ham}=3F_{ham}=3\cdot-6000=-18000\left(N\right)\)

\(F'_{ham}=ma'\Rightarrow a'=\dfrac{F'_{ham}}{m}=\dfrac{-18000}{2000}=-9\left(\dfrac{m}{s^2}\right)\)

\(v^2-v_0^2=2a'S'\)

\(\Leftrightarrow0^2-12^2=2\cdot-9\cdot S'\)

\(\Leftrightarrow-144=-18\cdot S'\)

\(\Rightarrow S'=8\left(m\right)\)

\(v_0=72\)km/h=20m/s

\(v=0\)m/s

Gia tốc vật: \(v^2-v^2_0=2aS\)

\(\Rightarrow a=\dfrac{v^2-v^2_0}{2S}=\dfrac{0-20^2}{2\cdot10}=-20\)m/s²

Lực cản:

\(F_c=m\cdot a=5000\cdot\left(-20\right)=-10^5N\)

Ta có: \(v^2-v_0^2=2as\)

\(\Rightarrow a=\dfrac{v^2-v_0^2}{2s}=\dfrac{0^2-10^2}{2\cdot50}=-1\left(m/s^2\right)\) 

Quãng đường mà vật di chuyển trong 4s kể từ lúc hãm phanh là:

\(s=v_0t+\dfrac{1}{2}at^2\)

\(\Rightarrow s=10\cdot4+\dfrac{1}{2}\cdot-1\cdot4^2=32\left(m\right)\)

\(v_0=72\)km/h=20m/s

Gia tốc vật:  \(v^2-v^2_0=aS\)

\(\Rightarrow a=\dfrac{v^2-v^2_0}{2S}=\dfrac{0-20^2}{2\cdot50}=-4\)m/s²

Lực hãm: \(F_c=-m\cdot a=-2\cdot1000\cdot\left(-4\right)=8000N\)

Thời gian đi đến lúc hãm phanh:   \(v=v_0+at\)

\(\Rightarrow t=\dfrac{v-v_0}{a}=\dfrac{0-20}{-4}=5s\)

\(v_0=72\)km/h=20m/s

Gia tốc xe: \(v^2-v^2_0=2aS\)

\(\Rightarrow a=\dfrac{v^2-v^2_0}{2S}=\dfrac{0-20^2}{2\cdot40}=-5\)m/s²

Ta có: \(-F_{hãm}-F_{ms}=m\cdot a\)

\(F_{hãm}=-F_{ms}-m\cdot a=-\mu mg-m\cdot a=-0,2\cdot1000\cdot10-1000\cdot\left(-5\right)=3000N\)