Task:
A 1.70 m wide, 1.40 m high, 1300 kg car hits a very slick patch of ice while going 20.0 m/s. Air resistance is not negligible. How long will it take until the car's speed drops to 16 m/s?
Solution:
For the car the drag force will be:
F d r a g = − 1 4 ⋅ w i d t h ⋅ h e i g h t ⋅ v 2 ⋅ k g m 3 F_{drag} = - \frac {1}{4} \cdot width \cdot height \cdot v ^ {2} \cdot \frac {kg}{m ^ {3}} F d r a g = − 4 1 ⋅ w i d t h ⋅ h e i g h t ⋅ v 2 ⋅ m 3 k g
Newton's 2 n d 2^{\mathrm{nd}} 2 nd
∑ i = 1 n F i = d p d t = M d v d t \sum_ {i = 1} ^ {n} F _ {i} = \frac {dp}{dt} = M \frac {dv}{dt} i = 1 ∑ n F i = d t d p = M d t d v ∑ i = 1 n F i = F d r a g \sum_ {i = 1} ^ {n} F _ {i} = F_{drag} i = 1 ∑ n F i = F d r a g M d v d t = − 1 4 ⋅ w i d t h ⋅ h e i g h t ⋅ v 2 ⋅ k g m 3 M \frac {dv}{dt} = - \frac {1}{4} \cdot width \cdot height \cdot v ^ {2} \cdot \frac {kg}{m ^ {3}} M d t d v = − 4 1 ⋅ w i d t h ⋅ h e i g h t ⋅ v 2 ⋅ m 3 k g d v v 2 = − 1 4 ⋅ M ⋅ w i d t h ⋅ h e i g h t ⋅ k g m 3 ⋅ d t \frac {dv}{v ^ {2}} = - \frac {1}{4 \cdot M} \cdot width \cdot height \cdot \frac {kg}{m ^ {3}} \cdot dt v 2 d v = − 4 ⋅ M 1 ⋅ w i d t h ⋅ h e i g h t ⋅ m 3 k g ⋅ d t ∫ 20 m s 16 m s d v v 2 = − ∫ 0 t 1 4 ⋅ M ⋅ w i d t h ⋅ h e i g h t ⋅ k g m 3 ⋅ d t \int_ {20 \frac {m}{s}} ^ {16 \frac {m}{s}} \frac {dv}{v ^ {2}} = - \int_ {0} ^ {t} \frac {1}{4 \cdot M} \cdot width \cdot height \cdot \frac {kg}{m ^ {3}} \cdot dt ∫ 20 s m 16 s m v 2 d v = − ∫ 0 t 4 ⋅ M 1 ⋅ w i d t h ⋅ h e i g h t ⋅ m 3 k g ⋅ d t − 1 v ∣ 20 m s 16 m s = − 1 4 ⋅ M ⋅ w i d t h ⋅ h e i g h t ⋅ t ⋅ k g m 3 ∣ 0 t - \frac {1}{v} \Big | _ {20 \frac {m}{s}} ^ {16 \frac {m}{s}} = - \frac {1}{4 \cdot M} \cdot width \cdot height \cdot t \cdot \frac {kg}{m ^ {3}} \Big | _ {0} ^ {t} − v 1 ∣ ∣ 20 s m 16 s m = − 4 ⋅ M 1 ⋅ w i d t h ⋅ h e i g h t ⋅ t ⋅ m 3 k g ∣ ∣ 0 t − 1 16 s m + 1 20 s m = − 1 4 ⋅ M ⋅ w i d t h ⋅ h e i g h t ⋅ t ⋅ k g m 3 - \frac {1}{16 \frac {s}{m}} + \frac {1}{20 \frac {s}{m}} = - \frac {1}{4 \cdot M} \cdot width \cdot height \cdot t \cdot \frac {kg}{m ^ {3}} − 16 m s 1 + 20 m s 1 = − 4 ⋅ M 1 ⋅ w i d t h ⋅ h e i g h t ⋅ t ⋅ m 3 k g − 1 80 s m = − 1 4 ⋅ 1300 k g m ⋅ 1.7 m m ⋅ 1.4 m m ⋅ k g m 3 - \frac {1}{80 \frac {s}{m}} = - \frac {1}{4 \cdot 1300 \frac {kg}{m}} \cdot 1.7 \frac {m}{m} \cdot 1.4 \frac {m}{m} \cdot \frac {kg}{m ^ {3}} − 80 m s 1 = − 4 ⋅ 1300 m k g 1 ⋅ 1.7 m m ⋅ 1.4 m m ⋅ m 3 k g t = 1 80 s m 1 4 ⋅ 1300 k g m ⋅ 1.7 m m ⋅ 1.4 m m ⋅ k g m 3 = 4 ⋅ 1300 80 ⋅ 1.7 ⋅ 1.4 s 1300 t = \frac {\frac {1}{80 \frac {s}{m}}}{\frac {1}{4 \cdot 1300 \frac {kg}{m}} \cdot 1.7 \frac {m}{m} \cdot 1.4 \frac {m}{m} \cdot \frac {kg}{m ^ {3}}} = \frac {4 \cdot 1300}{80 \cdot 1.7 \cdot 1.4} \frac {s}{\sqrt {1300}} t = 4 ⋅ 1300 m k g 1 ⋅ 1.7 m m ⋅ 1.4 m m ⋅ m 3 k g 80 m s 1 = 80 ⋅ 1.7 ⋅ 1.4 4 ⋅ 1300 1300 s
Answer:
t ≈ 27.311 s 1300 t \approx 27.311 \frac {s}{\sqrt {1300}} t ≈ 27.311 1300 s 
#339914 on Dec 2023