As soon as a traffic light turns green a car speeds up from rest to50.0 mi/h with a constant acceleration of 9.90 mi/h.s. in theadjoining bike lane, a cyclist speeds up from rest to 20.0 mi/hwith a constant acceleration of 13.5 mi/h.s. Each vehicle maintainsa constant velocity after reaching its cruising speed.

(a) For how long is the bicycle ahead of the car?
Answer in seconds
(b) By what maximum distance does the bicycle lead the car?
Answer in feet

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As soon as a traffic light turns green, a car speeds up from restto 46.0 mi/h with constant acceleration 8.50 mi/h-s. In theadjoining bike lane, a cyclist speeds up from rest to 24.0 mi/hwith constant acceleration 12.00 mi/h-s. Each vehicle maintainsconstant velocity after reaching its cruising speed.
(a) For what time interval is the bicycle ahead of the car?
(b) By what maximum distance does the bicycle lead the car?

As soon as a traffic light turns green, a car speeds up fromrest to50 mi/h with a constant acceleration of 9.0 mi/h per s. Intheadjoining bike lane, a cyclist speeds up from rest to 20 mi/hwith aconstant acceleration of 13.0 m/h per s. Each vehiclemaintains aconstant velocity after reaching its cruising speed. a)for how longis the bicycle ahead of the car? b)by what maximumdistance does thebicycle lead the car?

Here is what I got:

Car :

Initial velocity of car = 0m/s

Final velocity(vc) of car = 50mi/h

acceleration of car(ac) = 9.0 mi/h per s

Bicycle :

Initial velocity of bicycle = 0

Final velocity(vb) of bicycle = 20mi/h

Acceleration(ab) of bicycle = 13.0 m/h per s

Still I'm having problem solving this, and non of my classmatescould solve it either. No calcules please!

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

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