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

Solution:
The gap grows by 24 : 2 = 12 miles every hour.
Let the pedestrian's speed be x mph; then the cyclist's is 3x mph.
3x - x = 12, so 2x = 12 and x = 6.
Answer: pedestrian 6 mph, cyclist 18 mph.

Solution:
Distance from the hut: 5 . 3 = 15 miles.
Time for the way back: 15 : 3 = 5 hours.
Total time: 3 + 1 + 5 = 9 hours.
Answer: 9 hours.

Solution:
The two move apart at the sum of their speeds.
Speed of separation: 63 : 3 = 21 mph
Let the pedestrian's speed be x mph; then the cyclist's is 2.5x mph.
x + 2.5x = 21, so 3.5x = 21 and x = 6.
Answer: pedestrian 6 mph, cyclist 15 mph.

Solution:
By 9:00 the pedestrian has covered 6 . 1 = 6 miles.
The remaining 46 - 6 = 40 miles is covered while both are moving.
They move apart at 6 + 14 = 20 mph, so 40 : 20 = 2 hours are needed.
2 hours after 9:00 is 11:00.
Check: 3 . 6 + 2 . 14 = 18 + 28 = 46 miles.
Answer: 11:00.

Solution:
Together they cover 480 miles, and the first covers 80 miles more.
First car: (480 + 80) : 2 = 280 miles; second car: 480 - 280 = 200 miles.
Speeds: 280 : 4 = 70 mph and 200 : 4 = 50 mph.
Check: (70 + 50) . 4 = 480 miles.
Answer: 70 mph and 50 mph.

Solution:
In 1.5 hours the car covers 60 . 1.5 = 90 miles - this is the head start.
The motorcyclist gains 90 - 60 = 30 miles every hour.
Time to catch up: 90 : 30 = 3 hours.
In 3 hours the motorcyclist covers 90 . 3 = 270 miles, and this is the whole distance AB.
Check: the car travels 1.5 + 3 = 4.5 hours, 60 . 4.5 = 270 miles.
Answer: 270 miles.

Solution:
Let the speed be x mph. The distance is the same in both cases:
5x = 4(x + 15)
5x = 4x + 60
x = 60
Distance: 5 . 60 = 300 miles.
Check: 300 : (60 + 15) = 300 : 75 = 4 hours.
Answer: 60 mph and 300 miles.

Solution:
Downstream speed: 45 : 3 = 15 mph.
Upstream speed: 45 : 5 = 9 mph.
Own speed = (15 + 9) : 2 = 12 mph.
Speed of the current = (15 - 9) : 2 = 3 mph.
Check: 12 + 3 = 15 and 12 - 3 = 9.
Answer: own speed 12 mph, current 3 mph.

Solution:
Downstream speed: 36 : 2 = 18 mph.
Let the current be x mph; then the boat's own speed is 5x mph.
5x + x = 18, so 6x = 18 and x = 3.
Own speed: 5 . 3 = 15 mph; upstream speed: 15 - 3 = 12 mph.
Time upstream: 60 : 12 = 5 hours.
Answer: current 3 mph, own speed 15 mph, 5 hours.

Solution:
Let the speed of the second car be x mph; then the speed of the first car is (x + 14) mph.
In 4 hours the second car covers 4x miles and the first covers 4(x + 14) miles. Together that is the whole distance between them:
4x + 4(x + 14) = 520
4x + 4x + 56 = 520
8x = 464
x = 58
Second car: 58 mph; first car: 58 + 14 = 72 mph.
Check: (58 + 72) . 4 = 520 miles.
Answer: 72 mph and 58 mph.

Solution:
Take the whole distance as 1.
In one hour the freight train covers 1/6 of it and the fast train 1/4 of it.
Together in one hour: 1/6 + 1/4 = 2/12 + 3/12 = 5/12 of the distance.
Time until they meet: 1 : 5/12 = 12/5 = 2.4 hours.
Answer: 2 hours 24 minutes.

Solution:
The motorcycle's speed is 80 - 20 = 60 mph.
By 8:00 the car has covered 80 miles, so 220 - 80 = 140 miles are left.
They approach each other at 80 + 60 = 140 mph, so they need 140 : 140 = 1 hour.
They meet at 9:00; by then the car has travelled 80 . 2 = 160 miles.
Answer: at 9:00, 160 miles from the first town.

Solution:
Convert the speeds: 72 km/h = 20 m/s and 54 km/h = 15 m/s.
Relative speed: 20 - 15 = 5 m/s.
During the overtaking the fast train must gain the sum of the two lengths: 240 + 160 = 400 m.
Time: 400 : 5 = 80 s.
Answer: 80 seconds.

Solution:
Let the whole journey be 2d miles, so each half is d miles.
Time for the first half: d : 60 hours; time for the second half: d : 40 hours.
Total time: d/60 + d/40 = 2d/120 + 3d/120 = 5d/120 = d/24 hours.
Average speed = whole distance : whole time = 2d : (d/24) = 48 mph.
Note: the answer is not 50 mph, because the car spends more time on the slower half.
Answer: 48 mph.

Solution:
Going upstream the boat moves at 12 - 3 = 9 mph, while the raft drifts the other way at 3 mph, so they separate at 9 + 3 = 12 mph.
In 2 hours the gap becomes 12 . 2 = 24 miles.
After turning, the boat moves at 12 + 3 = 15 mph and the raft at 3 mph in the same direction, so the boat gains 15 - 3 = 12 miles each hour.
Time: 24 : 12 = 2 hours.
Note: relative to the water the raft does not move at all, and the boat always moves at 12 mph relative to the water - that is why the two times are equal.
Answer: 2 hours.

Solution:
Let the speed of the second car be x mph; then the speed of the first car is (x + 10) mph.
In 2 hours the second car covers 2x miles and the first covers 2(x + 10) miles. Together they cover the whole 300 miles:
2x + 2(x + 10) = 300
2x + 2x + 20 = 300
4x = 280
x = 70
Second car: 70 mph; first car: 70 + 10 = 80 mph.
Check: (70 + 80) . 2 = 300 miles.
Answer: 80 mph and 70 mph.

Solution:
Let the speed of the car be x mph.
By the moment the car catches up, the lorry has been travelling 3 + 2 = 5 hours and has covered 40 . 5 = 200 miles.
The car has been travelling 2 hours and has covered 2x miles. Both have covered the same distance:
2x = 40 . (3 + 2)
2x = 200
x = 100
Answer: 100 mph.

Solution:
Let x miles be the part he walked; then he cycled (45 - x) miles.
Time on foot: x/5 hours; time by bicycle: (45 - x)/15 hours. Altogether 5 hours:
x/5 + (45 - x)/15 = 5
Multiply both sides by 15:
3x + 45 - x = 75
2x = 30
x = 15
Check: 15 : 5 = 3 hours and 30 : 15 = 2 hours, 3 + 2 = 5 hours.
Answer: 15 miles.

Solution:
Let the boat's own speed be x mph.
Downstream speed: (x + 3) mph; upstream speed: (x - 3) mph.
The two distances are equal:
3(x + 3) = 5(x - 3)
3x + 9 = 5x - 15
9 + 15 = 5x - 3x
2x = 24
x = 12
Check: 3 . 15 = 45 miles and 5 . 9 = 45 miles.
Answer: 12 mph.

Solution:
There is no need to add up the fly's trips. The trains approach each other at [tex]50+50=100[/tex] mph, so they meet after [tex]200\div 100=2[/tex] hours. All this time the fly is flying:
[tex]75\times 2=150[/tex]

Solution:
During this time the front of the train travels the length of the bridge plus the length of the train:
[tex]480+120=600[/tex]
[tex]600\div 5=120[/tex]

Solution:
Every second the dog gets [tex]10-7=3[/tex] closer to the cat:
[tex]30\div 3=10[/tex]

Solution:
We do not need the distance between the cities. In the last [tex]10[/tex] minutes before meeting, the cars together cover exactly the distance between them, approaching at [tex]100+80=180[/tex] mph:
[tex]180\times \dfrac{10}{60}=30[/tex]

Solution:
Moving toward each other, the ants can be farther apart later only if they pass each other. In [tex]3[/tex] hours they crawl together [tex]4[/tex] (until they meet) plus [tex]8[/tex] (after that), i.e. [tex]12[/tex]. Per hour together:
[tex]12\div 3=4[/tex]
so each ant crawls [tex]4\div 2=2[/tex] ft per hour.

Solution:
Mitko walked until he was caught, i.e. [tex]20+2+3=25[/tex] minutes:
[tex]6\times \dfrac{25}{60}=\dfrac{5}{2}[/tex]
The car covered the same distance in [tex]3[/tex] minutes:
[tex]\dfrac{5}{2}\div \dfrac{3}{60}=\dfrac{5}{2}\times 20=50[/tex]

Solution:
Stoyan needs [tex]20+10=30[/tex] minutes less than Maria, and Dani only [tex]20-10=10[/tex] minutes less. Stoyan covers the same road in less time, so he is faster.

Solution:
Twice as fast on a path half as long means that the Wolf needed [tex]4[/tex] times less time. If her time is [tex]t[/tex] minutes, his time is [tex]\frac{t}{4}[/tex]:
[tex]t-\dfrac{t}{4}=30 \Rightarrow \dfrac{3t}{4}=30 \Rightarrow t=40[/tex]