8 - RTD
columns: r x t = d rows: objects traveling | Rate | × Time | = Distance | Object Traveling:
RTD often solved by using a Matrix
Convert the time-zone of the destination to the time zone of the origin to accurately compute the time traveled
RTD with time-zone changes
we must account for the difference b/w their speeds Let the slower objects speed be some variable r and the faster object's speed r+the difference in speeds
- when one object travels faster/slower than the other but we don't know the speed of either
when 2 objects leave at the same time and converge at a constant rate, they will have traveled for the same time at the instant they meet
-both objects leave at same time
their travel times must reflect the difference b/w their departure times to do this let the travel time of the object that leaves later be "t" and the object that leaves earlier be "t+ the difference b/w their departure times" that difference is then identical to the additional time experienced by the object that left earlier
-objects leave at different time and converge at a constant rate
When the 1st object is "x" times as fast as 2nd, let the rate of the 2nd object be "r"; 1st objects rate = xr When the 1st object is "x" % as fast as 2nd, let the rate of the 2nd object be "r"; 1st objects rate = (x/100) × r
-one object is relatively faster/slower than the other Some problems give us object's speed as a multiple of % of the other object's speed
Distance increasing b/w them at a rate of 70 mph ∴ 280/70 = 4 hrs
2 boats leave same place traveling opposite directions. rate A: 30 mph rate B: 40 mph How long b4 they are 280 miles apart
t for both is same i.e. 2 hrs Distance that A travels: r × 20 Distance that B travels: (r+10) × 20 → 20r + 20 We know the total distance since the distance that each both has traveled when they cross should sum to distance b/w ports 50mph
A & B leave opposite ports at the same time. One boat is traveling 10mph faster than the other. They cross each other at 2 hours. Distance between ports is 100 miles. Sum of both speeds ?
200/7 mph
Airplane's speed = 200 mph with wind it can travel 800 miles against same wind it can travel 600 in the same time What is the speed of the wind?
Average Rate = Avg. d / Avg t Avg Rate = (d₁ + d₂) / (t₁ + t₂)
Average Rate Questions - Formula
9000 minutes x 1 hr / 60 min → 900/6 → 150 hours 150/24 = 6¼ = 6 days + 1/4 day = 6 days + 6 hours Monday 11 AM + 6 hours = 5 PM NY Time, which is 5PM - 3 = 2 PM LA time
C left NY on Tuesday @ 11AM He drove to LA in 9,000 minutes. What day & time, in LA time, was it when he arrived in LA?
Distance A: 10 mpg | g | → 10g Distance B: 20 mpg | g | → 20g 10g + 20g = 300 g = 10 Distance A = 100 miles
Car A= 10 mpg Car B= 20 mpg Both driven such that each car consumed the same # of gallons of fuel. They traveled a combined distance of 300 miles. The car that was driven the fewest # of miles traveled how many miles?
Distance city: 20 mpg | g | → 20g Distance hwy: 30 mpg | 20-g | → 600-30g 500 = 20g + 600 - 30g → 100 = 10g → g = 10 Hwy miles: 600 - (30)(10) = 300 miles
Car A= 20 mpg in city ; 30 mpg on highway ON a certain day it was driven 500 miles in city & on highway & it consumed 20 gallons of gas. How many miles did it travel on highway?
Speed differential b/w the 2 objects allows the other to catch up When the 2 objects start from same point and one catches up to the other, both objects will have traveled the same distance when they meet
Catch-up Rate Problems ⊂⊃
Faster object's distance = + slower object's distance + any difference in starting points + any distance by which the faster object must pass the slower object
Catch-up and Pass Rate
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Catch-up and Pass Rate Scenarios: 1) faster object starts at the same location as the slowest & must catch up & pass the slower object and reach some distance beyond the slowest object 2) faster object starts at a location behind the slower object & must either catch up with the slower object or catch up with & pass the slower object by some distance
it is important to note that when this happens the total distance will be equal to the sum of individual distances each object travels
Converging Rate Questions → ← Objects moving towards each other
2 miles / 4 mph = .5 hours i.e. 30 mins for her to walk > ↑4mph she has to spend < ↓30 mins If she spends > ↑30 mins she is traveling < ↓4mph 1) 45 mins = .75 hrs - Sufficient 2) 120 mins / 60 = 2 hrs - Sufficient Question is asking if Rate S > 4 mph? looking at statement 2 we learn that her walking time is >2 hrs ∴ her rate is < 2miles / 2 hours i.e. < 1 mile/hour We want to know if Rate S > 4 mph, we know from (2) that her rate is < 1mph ∴ we know it is not > 4mph
Did S walks faster than 4 mph while walking the 2 miles, at a constant rate, from her home to park? 1) it took S longer than 45 minutes to walk 1) it took S longer than 120 minutes to walk
1) IS because we don't know how long he drove 2) IS because we don't know how fast he drove 1+2) Suff: suppose he drove 50 mph & for 2 hrs that would give 100 miles But he drove faster and took less time suppose 51 miles 1 hour or 100 mile in hour --- too many possibilities (E)
Did T drive more than 100 miles on his trip? 1) T's average rate for the trip was greater than 50 mph 2) The trip took T less than 2 hours
if 2 objects start at the same place & move in opposite directions, then the sum of the distances the 2 objects travel must equal the total distance traveled b/w them
Diverging Rate Questions ← →
C
Dog goes out and comes back same route: total distance 100 yards What is his avg rate? 1) on the way out constant rate of 40 yard/min 2) on the way back constant rate of 30 yards/min
40mph
H & Q traveling towards each other. Distance b/w each other 270m. Q's speed is 1/4 greater than H's. They meet after 3 hours. What is H's speed?
takes 30 mins to swim did travel more than 1.6 km 1) 1, 2, 3... km/ hr IS 2) > 3 km / hour i.e in half an hour > 1.5 km/hr 1.51, 1.52... IS 1+2) IS (E)
If it takes a fish 30 minutes to swim straight from A to B, did the fish swim more than 1 mile (1m = 1.6km) 1) fish travels at a constant rate that is less than 4 km/hr 2) fish travels at a constant rate that is greater than 3 km/hr
If you have no info on speed, you can state actual speed as r and the hypothetical speed as (r+x mph)
If/Then Rate Questions
If Jean had traveled at 55mph, she would have arrived 2 hours later than she did If Paula had swum 1 mph faster, she would've completed the race 45 seconds sooner If A flew her plane 50 mph slower, she would've arrived 2 hour later
If/Then Rate Questions provide us with info about a scenario that actually happened, and info about a scenario that would have happened if the set of conditions had been different Eg: if [object] had traveled [some rate], it would have [saved/added] t hour to its time
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Inverse Proportionality - rate is inversely proportional to the time P take "more than" 20 minutes to walk 1 mile, since rate ∞1/t, ∴ her rate is less than 1 mile per 20 minutes, i.e. r < ¹/₂₀ mile/minute If her time was "exactly" 20 minutes, her rate would be ¹/₂₀ mile/minute.
h→s 40t s→h 60(2-t) = 120-60t since distance each way is same 40t=120-60t t = 1.2 hours d= 40(1.2) = 48 miles Total distance is 2d ∴ 2 × 48 = 96
Jack home to school @ 40mph school to home @ 60mph Total time = 2 hours Total distance traveled?
Suppose distance is "d" t₁ = d/4 t₂= d/10 Total distance = 2d Rate = 2d / (d/4 +d/10) → 2d / [(5d + 2d)/20] →2d/(7d/20) →(2d) × 20/7d → 40/7 mph Or you can pick a # for distance - pick 20 t₁ = 20/4 = 5 t₂= 20/10 = 2 Total distance = 40 ∴40/7 mph
M walked from home to school at 4mph, ran back to home at 10mph. Average rate?
1) P: | 2r | × | t | = → 2rt S: | r | × | t | = → rt The distance should sum to 2 2rt +rt = 2 → 3rt = 2 → rt =²/₃ → Insufficient since we don't know "r" 2) we only know r & have no info on P's rate - IS Combined we now have r and we can calculate P's rate - Sufficient (C)
P & S leave same place traveling opposite directions. How long b4 they are 2 miles apart? 1) P's rate is 2x S's 2) If S ran the entire 2 miles on his own, he would've completed in 20 mins
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Proportionality ---- RTD → r = d/t d∞ r & t | r ∞ d | r ∞ 1/t | t ∞ 1/r & d If rate increases & time remains constant, then distance must increase If time increases & rate remains constant, then distance must increase
a ratio that compares 2 quantities of 2 different units of measurement Miles/hour Calories/mile Chocolate bars/week Gallons/second
Rate
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Rate in Miles per Gallon (mpg) - Efficiency You can make the same matrix as mph | r | x | gallons | = d
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Relative Motion Rate Boat traveling at 10 mph in still water, if going with current which is at 4 mph the boat will travel at 10 mph +4 mph → 14 mph
In round trip problems when only total travel time is provided, consider letting the time to a destination equal some variable t & the time back equal "total trip time - t"
Round Trip Problems ⊂⊃
times for both will be same ∴ t for S = d/800 t for J = (d+80+40)/1200 set equations equal: d/800 = (d+120)/1200 (12/8) d = d+120 → 3d/2 = (d+120) → 3d = 2d +240 → d = 240 ∴ t using equation 1 = 240/800 = 3/10 mins → 18 seconds
S & J are running on a track. J is 80 feet behind rate S: 800 feet/min rate J: 1200 feet/min How many seconds b4 J catches up and gets 40 feet ahead?
1) r> 6mph ∴ time < 3/6 i.e. .5 hours --- IS without knowing R's info 2) R rate is 2 times S, we don't know the "r" or the "t" -- -IS 1+2) if it takes S 0.5 hr to run 3 miles & R is 2 times faster, then he must finish the same distance in half the time i.e. .25hrs x+y = 45 mins (C)
S & R run on a 3 mile loop. R takes x minutes & S takes y mins, is x + y > 60 mins? 1) S jogs at a constant rate that is greater than 6mph 2) R runs twice as fast as S
rt =d → r = d/t
Speed Formula
144 minutes They both will have traveled same time when they meet | Rate | × Time | = Distance | T: | 60 | t →| 60t E: | 40 | t → | 40t d₁+d₂ = 240 60t + 40t = 240 → 2.4 hrs x 60 → 144 mins Alternate : both closing distance at 100 mph ∴ 240miles /100mph = 2.4 hr x 60min/1hr = 24x6 = 144
T & E are 240 miles apart. They both travel towards each other; T @ 60mph, E @ 40mph. How many minutes b4 they meet?
Their distance is the same ∴ Distance T: 8t Distance S: 12(t-.5) → 12t - 6 8t = 12t - 6 → t = ³/₂ 1.5 hours i.e since T started running it has been 1.5 hours but S started an 30 mins later ∴ it will take S 1.5 - .5 hours to catch up i.e. 1 hour, which makes sense since she is faster; ran the same distance in 1 hour which T ran in 1.5 hrs
T starts running on a track. S starts running after 30 mins. rate T: 8 mph rate S: 12 mph How long b4 S catches up?
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There is an important relationship b/w the numerator and the denominator. If you say something is moving at a constant rate of 4miles/hr, you are saying if 4 miles elapse, 1 hour elapses & vica versa
Faster Time + 1 = Slower Time Actual: | r | . | slower time | = 60 Hypot: | r + 2 | . | faster time | = 60
Thomas riding at constant rate. Distance to store 60 miles. If he had ridden 2 mph faster than he actually did, he would have saved 1 hr. How fast did he actually ride to the store?
2:30 PM t for the one that left later t+0.5 for the earlier train | Rate | × Time | = Distance | T: | 50 | t +0.5 →| 50t + 25 E: | 60 | t → | 60t 245 = 50t + 25 + 60t → 220 = 110t → 2 hours ∴ E left at 12:30 add 2 hours → 2:30 PM
Trains T & E are 245 miles apart. They both travel towards each other; T starts at 12:00 PM @ 50mph, E starts at 12:30 PM @ 60mph. At what time do they pass each other?
this can be solved by understanding that r ∞ d/t Ratio of their speeds: 2:1 Ratio of distance they cover 2:1 ∴ V has covered 2/3 of the distance in 10 mins He will take another ²/₃x = 10 x = 30/2 = 15 mintues C
V & S traveling towards each other's houses. How long will V take to get to S's house? 1) V's speed is 2 times S's 2) They pass each other at 10 minutes
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Variations -both objects leave at same time (when they meet t is same for both) -objects leave at different time (t & t+difference) -one object travels faster/slower than the other (r & r+difference) -one object is relatively faster/slower than the other (r & x.r) (r & x/100 . r)
d₁ = distance object 1 d₂ = distance object 2 d₁ + d₂ = d total
When 2 objects converge at a constant rate and meet, the total distance the objects cover is defined by the equation
