Motion Problems - Distance, Speed, Time: Very Difficult Problems with Solutions

Solution:
The car left the factory at the usual time and drove the usual way, but turned back at the meeting point. The [tex]12[/tex] minutes it saved are the trip from the meeting point to the bus stop and back, i.e. [tex]6[/tex] minutes each way. So the car met the director [tex]6[/tex] minutes before it would have reached the stop at [tex]8{:}00[/tex], i.e. at [tex]7{:}54[/tex].

Solution:
Let the man's speed be [tex]u[/tex], the tractor's [tex]v[/tex] and the length [tex]L[/tex]. Going forward he moves relative to the pipe at [tex]u-v[/tex] for the time [tex]\frac{150}{u}[/tex]; going back at [tex]u+v[/tex] for the time [tex]\frac{20}{u}[/tex]:
[tex]L=(u-v)\times \dfrac{150}{u}=(u+v)\times \dfrac{20}{u}[/tex]
[tex]150(u-v)=20(u+v) \Rightarrow 130u=170v \Rightarrow v=\dfrac{13}{17}u[/tex]
[tex]L=150\left(1-\dfrac{13}{17}\right)=150\times \dfrac{4}{17}=\dfrac{600}{17}=35\dfrac{5}{17}[/tex]

Solution:
The bicycle can carry someone at most [tex]12[/tex] miles in total, so one of the two rides at most [tex]6[/tex] miles and walks at least [tex]6[/tex] miles. That person needs at least
[tex]\dfrac{6}{20}+\dfrac{6}{5}=0.3+1.2=1.5[/tex]
hours, i.e. [tex]90[/tex] minutes. This time is achieved: the first rides [tex]6[/tex] miles, leaves the bicycle and walks on; the second walks [tex]6[/tex] miles (in [tex]1.2[/tex] h, when the bicycle is already there) and rides the rest. Both arrive after [tex]90[/tex] minutes.

Solution:
Observe the motion from truck A (as if A stood still). Then B and C move forward with some speeds [tex]b[/tex] and [tex]c[/tex] relative to A. At first both are [tex]18[/tex] behind A. While C covers these [tex]18[/tex], B covers only [tex]18-13.5=4.5[/tex], so
[tex]\dfrac{b}{c}=\dfrac{4.5}{18}=\dfrac{1}{4}[/tex]
B reaches A after covering [tex]18[/tex]; in the same time C covers [tex]4\times 18=72[/tex], so it is [tex]72-18=54[/tex] miles ahead of A.

Solution:
Let the truck's speed be [tex]v[/tex]. The motorcyclist catches the truck [tex]t[/tex] hours after his start:
[tex]50t=v\left(t+\dfrac{1}{2}\right) \Rightarrow t=\dfrac{v}{2(50-v)}[/tex]
He rides back the same distance in the same time [tex]t[/tex], so he is back at A [tex]\frac{1}{2}+2t[/tex] hours after the truck started, which is the truck's travel time:
[tex]\dfrac{1}{2}+\dfrac{v}{50-v}=\dfrac{180}{v}\ \Big|\times 2v(50-v)[/tex]
[tex]v(50-v)+2v^2=360(50-v) \Rightarrow v^2+410v-18000=0[/tex]
[tex](v-40)(v+450)=0 \Rightarrow v=40[/tex]
Check: the motorcyclist catches the truck after [tex]2[/tex] h, [tex]100[/tex] miles from A, and is back at A [tex]4.5[/tex] h after the truck started; the truck needs [tex]180\div 40=4.5[/tex] h.

Solution:
Let sunrise be [tex]x[/tex] hours before noon and the speeds [tex]a[/tex] (from A) and [tex]b[/tex] (from B). The part of the road covered by the first woman before noon is covered by the second after noon in [tex]9[/tex] hours, and vice versa (in [tex]4[/tex] hours):
[tex]ax=9b[/tex], [tex]bx=4a[/tex]
[tex]ab\times x^2=36ab \Rightarrow x^2=36 \Rightarrow x=6[/tex]
Sunrise was at 6 a.m.

Solution:
Let the speeds be [tex]a[/tex] (motorcyclist) and [tex]b[/tex] (cyclist). Until the meeting, in [tex]\frac{7}{6}[/tex] h:
[tex]\dfrac{7}{6}(a+b)=84 \Rightarrow a+b=72[/tex]
After the meeting the motorcyclist rides the cyclist's part [tex]\frac{7}{6}b[/tex] to B, rests [tex]\frac{1}{3}[/tex] h and rides back for the rest of the hour. One hour after the meeting both are at the same point; its distance from B is:
[tex]a\left(1-\dfrac{1}{3}\right)-\dfrac{7}{6}b=\dfrac{7}{6}b+b \Rightarrow \dfrac{2}{3}a=\dfrac{10}{3}b \Rightarrow a=5b[/tex]
[tex]6b=72 \Rightarrow b=12[/tex], [tex]a=60[/tex]

Solution:
Let [tex]L[/tex] be the length and [tex]v[/tex] the speed of the trains. When overtaking, the train passes him with relative speed [tex]v-5[/tex]; when meeting him, with [tex]v+5[/tex]:
[tex]L=9(v-5)=7(v+5)[/tex]
[tex]9v-45=7v+35 \Rightarrow v=40[/tex], [tex]L=9\times 35=315[/tex]