Q: As soon as a traffic light turns green, a car speeds up fromrestto 50 miles/hours with constant acceleration 9miles/hour*seconds.In the adjoining bike lane, a cyclist speeds upfrom rest to 20miles/hour with constant acceleration 13miles/hours*seconds. Eachvehicle maintains constant velocity afterreaching its cruisingspeed.
a) For what time interval is the bicycle ahead of the car?
b) By what maximum distance does the bicycle lead the car?
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In step 3, I do not understand where the second equation camefrombecause it is not given in the text. Can you explain how youcomeup with that equation?
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Step 2 includes the calculations in finding the time for the biketoreach the cruising speed which is 1.54 seconds.
time = (final velocity - initial velocity)/ acceleration
time = (20-0)/13
time = 1.54 seconds
Step 3 says "For the car has to overcome the bicycle both hastotravel same distance in the same time 't'.
1st equation: distance = (initial velocity*time) +(.5*accelerationof car*time squared)
distance = 0 + (.5*9mi/hs*timesquared)
distance = 4.5*time squared
2nd equation:
distance = (initial velocity of bike*time) + (.5*accelerationofcar*time of bike squared) + (final velocity of car)*(time- timeofbike)
distance = 0 + (.5*13 mi/hs*1.54seconds squared)+(20mi/h)*(t-1.54s)
distance = 20t-15.4
-- afterwards it says to take both equations and equal them toeachother to solve for time
which gives the answer where the bicycle is ahead of the car int(time) = 3.45 seconds
Q: As soon as a traffic light turns green, a car speeds up fromrestto 50 miles/hours with constant acceleration 9miles/hour*seconds.In the adjoining bike lane, a cyclist speeds upfrom rest to 20miles/hour with constant acceleration 13miles/hours*seconds. Eachvehicle maintains constant velocity afterreaching its cruisingspeed.
a) For what time interval is the bicycle ahead of the car?
b) By what maximum distance does the bicycle lead the car?
----------------
In step 3, I do not understand where the second equation camefrombecause it is not given in the text. Can you explain how youcomeup with that equation?
----------------
Step 2 includes the calculations in finding the time for the biketoreach the cruising speed which is 1.54 seconds.
time = (final velocity - initial velocity)/ acceleration
time = (20-0)/13
time = 1.54 seconds
Step 3 says "For the car has to overcome the bicycle both hastotravel same distance in the same time 't'.
1st equation: distance = (initial velocity*time) +(.5*accelerationof car*time squared)
distance = 0 + (.5*9mi/hs*timesquared)
distance = 4.5*time squared
2nd equation:
distance = (initial velocity of bike*time) + (.5*accelerationofcar*time of bike squared) + (final velocity of car)*(time- timeofbike)
distance = 0 + (.5*13 mi/hs*1.54seconds squared)+(20mi/h)*(t-1.54s)
distance = 20t-15.4
-- afterwards it says to take both equations and equal them toeachother to solve for time
which gives the answer where the bicycle is ahead of the car int(time) = 3.45 seconds
