How Far Can a Bicycle Lead a Car When Both Accelerate from a Stop?

[feihong47]

As soon as a traffic light turns green, a car speeds up from rest to 50.0 mi/h with constant acceleration 9.00 mi/h/s. In the adjoining bicycle lane, a cyclist speeds up from rest to 20.0 mi/h with constant acceleration 13.0 mi/h/s. Each vehicle maintains constant 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?after conversion of units to m/s:
car a=4.022, v=22.35
bike a =5.81, v = 8.94

car reaches max v in 22.35/4.022 = 5.56 s
car displacement (if t < 5.56) is 0.5a_cart², and (t > 5.56) 0.5a_car(5.56)² + 22.35(t-5.56)

bike reaches max v in 8.94/5.81 = 1.54 s
bike displacement (if t < 1.54) is 0.5a_biket², and (t > 1.54) 0.5a_bike(1.54)² + 8.94(t-1.54)

at t=1.54, bike is ahead of car
at t=5.56, car is ahead of bike
so for part a) i know 1.54 < t < 5.56

so equation is 0.5a_cart² = 0.5a_bike(1.54)₂ + 8.94(t-1.54)

solving, t = 3.45 smy question is how to get part B? From a prev question, they said something about setting the velocity equal.

so I set bike 8.94 = a_cart

t = 2.22

putting it in car equation x = 9.91, and bike equation x = 12.97

So their greatest separation would be about 3.06 m. IS THIS CORRECT? I don't have an answer key and someone else's solution does not match mine.

thanks!

 

[PeterO]

As soon as a traffic light turns green, a car speeds up from rest to 50.0 mi/h with constant acceleration 9.00 mi/h/s. In the adjoining bicycle lane, a cyclist speeds up from rest to 20.0 mi/h with constant acceleration 13.0 mi/h/s. Each vehicle maintains constant 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?


after conversion of units to m/s:
car a=4.022, v=22.35
bike a =5.81, v = 8.94

car reaches max v in 22.35/4.022 = 5.56 s
car displacement (if t < 5.56) is 0.5a_cart², and (t > 5.56) 0.5a_car(5.56)² + 22.35(t-5.56)

bike reaches max v in 8.94/5.81 = 1.54 s
bike displacement (if t < 1.54) is 0.5a_biket², and (t > 1.54) 0.5a_bike(1.54)² + 8.94(t-1.54)

at t=1.54, bike is ahead of car
at t=5.56, car is ahead of bike
so for part a) i know 1.54 < t < 5.56

so equation is 0.5a_cart² = 0.5a_bike(1.54)₂ + 8.94(t-1.54)

solving, t = 3.45 s


my question is how to get part B? From a prev question, they said something about setting the velocity equal.

so I set bike 8.94 = a_cart

t = 2.22

putting it in car equation x = 9.91, and bike equation x = 12.97

So their greatest separation would be about 3.06 m. IS THIS CORRECT? I don't have an answer key and someone else's solution does not match mine.

thanks!

Your reasoning on part B is very good. If your arithmetic throughout is as good there is no reason for your answer to me wrong [I calculate your answer when using your velocity and mph conversions - which i didn't check.