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

Solution:
Let the distance be [tex]d[/tex], the boat's speed in still water [tex]v[/tex] and the speed of the current [tex]c[/tex]:
[tex]v+c=\dfrac{d}{2}[/tex], [tex]v-c=\dfrac{d}{3}[/tex]
[tex]2c=\dfrac{d}{2}-\dfrac{d}{3}=\dfrac{d}{6} \Rightarrow c=\dfrac{d}{12}[/tex]
The raft moves with the speed of the current, so it needs [tex]d\div \frac{d}{12}=12[/tex] hours.

Solution:
The ring floats with the current; in [tex]5[/tex] hours it covers [tex]10[/tex], so the current is [tex]c=2[/tex]. If [tex]d[/tex] is the distance and [tex]v[/tex] the boat's own speed:
[tex]\dfrac{d}{5}-\dfrac{d}{10}=(v+c)-(v-c)=2c=4[/tex]
[tex]\dfrac{d}{10}=4 \Rightarrow d=40[/tex]

Solution:
Look at the motion relative to the water (as if the water were still). Then the hat stays in place, and the boat moves away from it for [tex]10[/tex] minutes and comes back with the same speed [tex]12[/tex], i.e. also in [tex]10[/tex] minutes. So the hat is caught [tex]20[/tex] minutes after it fell. During this time it floated with the current:
[tex]3\times \dfrac{20}{60}=1[/tex]

Solution:
Let [tex]L[/tex] be the length, [tex]e[/tex] the speed of the escalator and [tex]u[/tex] the man's own speed:
[tex]L=30(e+u)=20(e+3u)[/tex]
[tex]30e+30u=20e+60u \Rightarrow 10e=30u \Rightarrow e=3u[/tex]
[tex]L=30(3u+u)=120u[/tex], [tex]\dfrac{L}{e}=\dfrac{120u}{3u}=40[/tex]

Solution:
Let [tex]AB=d[/tex]. Until the first meeting the two cars together cover one distance [tex]d[/tex]; until the second meeting they cover three distances [tex]3d[/tex]. With constant speeds, each car covers [tex]3[/tex] times more by the second meeting than by the first. The car from A covered [tex]70[/tex] before the first meeting, so by the second meeting it covered [tex]3\times 70=210[/tex], which is [tex]d+40[/tex]:
[tex]d+40=210 \Rightarrow d=170[/tex]

Solution:
Let the speeds be [tex]a[/tex] (car) and [tex]b[/tex] (truck) and [tex]AB=D[/tex]; measure positions from A. After [tex]1[/tex] hour the car is at [tex]a[/tex] and the truck at [tex]D-b[/tex]:
[tex]a=\dfrac{0+(D-b)}{2} \Rightarrow D=2a+b[/tex]
After [tex]t[/tex] hours the car is at [tex]at[/tex], the truck at [tex]D-bt[/tex], and the midpoint between the truck and B is at [tex]\frac{(D-bt)+D}{2}=D-\frac{bt}{2}[/tex]:
[tex]at=D-\dfrac{bt}{2} \Rightarrow t\left(a+\dfrac{b}{2}\right)=2a+b \Rightarrow t=2[/tex]
So this happens at [tex]14{:}00[/tex] (2 p.m.).

Solution:
Riding in opposite directions, they meet each time they have covered together one full lap:
[tex]400\div (10+15)=16[/tex]
[tex]7\times 16=112[/tex]

Solution:
Every day Friday gets [tex]33-23=10[/tex] further ahead. They are together again when he is one full lap ([tex]100[/tex]) ahead, i.e. after [tex]10[/tex] days. By then Robinson has walked
[tex]23\times 10=230=2\times 100+30[/tex]
i.e. two full laps and [tex]30[/tex] more. They meet [tex]30[/tex] miles from the boat.

Solution:
Let him pass C [tex]T[/tex] hours after [tex]10{:}30[/tex]. On the first day [tex]AC=8T[/tex]. On the second day he started [tex]2.5[/tex] hours later in the day, so [tex]BC=10(T-2.5)[/tex]:
[tex]8T+10(T-2.5)=38 \Rightarrow 18T=63 \Rightarrow T=3.5[/tex]
He passed C at [tex]10{:}30[/tex] + [tex]3[/tex] h [tex]30[/tex] min = [tex]14{:}00[/tex]. (Check: [tex]AC=28[/tex], [tex]BC=10[/tex].)

Solution:
The motorcyclist rides at [tex]2.5\times 20=50[/tex] and needs [tex]60\div 50=1.2[/tex] h. In this last [tex]1.2[/tex] h the cyclist covers [tex]20\times 1.2=24[/tex], so when the motorcyclist started, the cyclist was
[tex]60-24=36[/tex]
miles from A.

Solution:
Until the catch-up Peter ran from [tex]1200[/tex] yd to [tex]4800[/tex] yd, i.e. [tex]3600[/tex] yd, which took [tex]3600\div 180=20[/tex] min. In these [tex]20[/tex] min Ivan ran [tex]4800[/tex] yd, so his speed is
[tex]4800\div 20=240[/tex] yd/min
Ivan runs the last [tex]200[/tex] yd in [tex]\frac{200}{240}=\frac{5}{6}[/tex] min; meanwhile Peter runs [tex]180\times \frac{5}{6}=150[/tex] yd and is [tex]200-150=50[/tex] yd from the finish.

Solution:
Let his own speed be [tex]u[/tex] and the speed of the current [tex]c[/tex] (per minute). Upstream he moved backward:
[tex]4(c-u)=80 \Rightarrow c=u+20[/tex]
After turning, he drifted [tex]3[/tex] minutes with the current and paddled [tex]2[/tex] minutes downstream:
[tex]3c+2(u+c)=240 \Rightarrow 5c+2u=240[/tex]
[tex]5(u+20)+2u=240 \Rightarrow 7u=140 \Rightarrow u=20[/tex]

Solution:
Let the distances be [tex]x[/tex], [tex]45x[/tex], [tex]4x[/tex] and the times [tex]4y[/tex], [tex]20y[/tex], [tex]y[/tex]:
[tex]\dfrac{x+45x+4x}{4y+20y+y}=\dfrac{50x}{25y}=\dfrac{2x}{y}=32 \Rightarrow \dfrac{x}{y}=16[/tex]
Walking speed:
[tex]\dfrac{x}{4y}=\dfrac{16}{4}=4[/tex]

Solution:
Measure everything in mouse steps. The cat is [tex]5\times 10=50[/tex] steps behind the mouse. During one jump the cat covers [tex]10[/tex] and the mouse [tex]3[/tex], so the cat gains [tex]7[/tex] steps per jump and needs [tex]\frac{50}{7}=7\frac{1}{7}[/tex] jumps to catch up.
The mouse reaches its hole in the time of [tex]\frac{20}{3}=6\frac{2}{3}[/tex] jumps, i.e. earlier. The mouse escapes.