13) Rates Flashcards
Rate A is “1/4 greater than” rate B?
NOT sum…
Means 25% greater or *5/4
REVISIT WORDING IN math special
What is a rate?
a rate COMPARES TWO different UNITS OF MEASUREMENT
e.g., miles per hour, chocolate bars per week, seconds per minute
One of the most common types of rate problems: Rate-Distance-Time problem = “speed problem”
Speed problems
Helpful like all word problems to draw a table (matrix) to keep track of inputs:
Columns: Rate * time = Distance
Rows: each object
Make sure units are all compatible first before solving!
Speed problem variation #1) elementary rate questions
Test basic understanding of the rate equation
Rate * time = distance
Speed problem variation #2) average rate questions
Dealing with non-constant rates over a period of time
> can STILL use rate-time-distance formula
Average rate = TOTAL distance / TOTAL time
= (average rate1)(t1 / total t) + (average rate2)(t2 / total t)
> NOT (average rate 1 + average rate 2)/2 —> because of unequal weighting of times (different lengths of time at which the object may have traveled at each rate)
e.g., what if a car travelled 10 hours for 60 miles per hour and 1 minute at 70 miles per hour
Watchouts:
> Write down ALL conditions (even hidden ones) e.g., distances are EQUAL for a round trip
> average rate problems often involve other quirks (e.g., round trip, catch up, converging)
Melissa walked from home to school at an average rate of 4mph. She realized when she got to school that she had forgotten her math book, so she ran home at an average rate of 10 mph to get it. What was her average speed for this round trip?
THIS IS SUFFICIENT if we let d = distance of one leg –> get cancelled out
We don’t know t —> express using rates and d
Average Rate = Total D / Total Time
Time for leg 1 = D / 4
Time for leg 2 = D / 10
Therefore:
Average Rate = 2D / (D / 4 + D / 10)
= 5 5/7 mph
Alternatively, we know 4T1 = 10T2 —> could sub one variable into average rate equation and let variable cancel out
Speed problem variation #3) Converging rate question
Two objects start at opposite ends and MOVE TOWARD EACH OTHER
Distance Travelled by Object 1 + Distance Travelled by Object 2 = Total Distance between them ——-> when they meet
Be mindful of TIME
Usually dealing with CONSTANT RATES
Sub-variations (in converging rate Q but also any rate Q)
> objects leave at the same time —> T1 = T2 = T when they meet
> objects leave at different times —-> adjust travel times or distances
e.g., T1 = T2 + 5 minutes (object 1 had 5 minutes more time to travel than object 2) —-> earlier object’s time = later object’s time + difference
let t be time when both objects are travelling together
Alternatively: Distance travelled by object 1 earlier + distance travelled by object 1 + distance travelled by object 2 = total distance
> one object travels faster or slower than the other object —-> represent slower object with r and faster object r + difference in speeds
e.g., A’s speed is 10mph faster than B’s speed
> one object is relatively faster or slower than the other object (%, multiple, fraction) ——> represent object’s speed as a FACTOR (*fraction)
Speed problem variation #4) Diverging rate question
Two objects move AWAY from each other, creating a gap
Distance Travelled by Object 1 + Distance Travelled by Object 2 + initial gap = Total Distance between them
OR gap between objects GROWS at a combined rate
Same variations as converging rate Q applied
> initial gap (objects leave at different times)
Speed problem variation #5) Round-trip rate question
An object is travelling to some destination along a path, THEN returning to the starting point
2 * distance of each path = Total Distance
Distance travelled by one path = Distance travelled by another path —> super important if not given distances
BE MINDFUL OF TIME (time it takes to go to a destination can differ from the time it takes to return due to different rates)
> Total Time to complete round trip = t for one leg + t for another leg
e.g., J drove from home to school in the morning at an average rate of 40 mph. In the evening, he drove back home from school at an average rate of 60 mph. If he spent a total of 2 hours traveling and he took the same roads each way, what is the total distance J travelled?
Let x = distance of one route
Total distance = 2x
MATRIX:
Route 1: 40 * t1 = x
Route 2: 60 * (2 - t1) = x
Need to know t1 to know total distance —-> x = x
40t1 = 60 * (2 - t1)
t1 = 6/5 hours
therefore, 2x = 2 * 40 * 6/5 = 96 m
Speed problem variation #6) Catch-up rate question
Two objects moving in the SAME direction with some initial gap, and eventually the faster object catches up to the slower object
> includes Circular track qs
Initial Gap + Distance Travelled by Slower Object during time = Distance Travelled by Faster Object during time
Usually involves constant speeds
OR gap closes at rate = faster rate - slower rate
Keep in mind times
Other important variations:
> Faster object catches up with slower object AND passes slower object to reach some distance beyond it (Catch up and pass)
Initial gap + distance travelled by slower object during time + new gap = distance travelled by faster object during time
OR time to catch up to slower object + time to create new gap
Gap approach in solving rate questions
CONVERGING OBJECTS (opposite direction): Rate at which gap closes = SUM of individual rates (both are contributing to the gap closure)
DIVERGING OBJECTS (opposite direction): Rate at which gap expands = SUM of individual rates
CATCH UP OBJECTS (same direction): Rate at which gap closes = Fast rate - slow rate
Helpful formula:
Change Rate * time = Change Distance
Why do we need to learn this?
> can save you TIME
Speed problem variation #7) Relative motion rate question
Recall physics: Outside force acting on an object can be POSITIVE or NEGATIVE
Example scenarios: flying with wind, sailing with a current
We care about the RELATIVE SPEED (on the ground) –> actual rate of the object isn’t going faster
Boat’s speed + current speed = relative speed
Boat’s speed - current speed = relative speed
> use answer choices to help guide you (if you have d = d and r for boat’s speed and c for current’s speed, you can derive a RATIO and use answer choices in Two Part Analysis to help)
Circular track considerations
If the faster object and slower object start at the same time with no gap, then faster object will COMPLETE 1 LAP around (equal to the distance of the circle), AND travel the same distance of the slower object
Distance of slower object + total distance of lap = Distance of the faster object
Speed problem variation #8) If-then rate questions
“If [object] had traveled [some rate faster or slower], it would have [saved/added] t hours to its time”
“If [object] had traveled [at some rate], it would have [saved/added] t hours to its time”
SAME DISTANCE travelled
Solve by:
> setting up variables for ACTUAL SPEED and ACTUAL TIME
> substituting one variable into the other equation and solving
All about keeping track of:
> Actual trip
> Hypothetical trip
Likely end up with quadratic equation that you can solve (goal is to end up with 1 variable in the equation)
Be careful of time-zone changes in rate-time-distance problems
EST vs Mountain time —-> need to convert to the SAME TIME ZONE
Will be given time zone conversion rate
