There do not seem to be any accelerations going on here.
First cyclist:
how long will he take"
outbound:
20 km / 20km/h = 1 h
inbound:
20 km / 15 km/hr = (4/3) h
total time for C1 = (7/3) hour
so
Second cyclist must do 2x km at 10 km/hr in (7/3) hr
2 x = 10 km/hr * (7/3) hr
x = 35/3 km
= 11 2/3 km out and the same back
Using the following equations as a guide (some may be used, some may not-choose wisely)
y1 OR x1= initial position
x2 or Y2 =final position
V1 or Vi= Initial velocity
V2 or Vf- Final velocity
a= Acceleration
T- time
Equation 1.) x2=x1 +vt
Equation 2.) v2= v1+at
Equation 3.) AvgV=(v1+v2)/2
*acceleration must be constant
Equation 4.) y2=y1 +V1t +1/2 at^2
Equation 5.) V2^2=v1^2 +2a (x2-x1)
*acceleration must be constant
1.) Two cyclists start out on the same trail together that is
overall 20km long. The first cyclist completes the entire
trail and is traveling at a constant 20 km/hr for the away
part of the trip and then only 15 km/hr for the return trip.
the Second cyclist, who is slower, is moving along at a
Constant 10 km/hr for the entire trip. The second cyclist
wants to turn around at some point so that he will arrive
back at the starting position at the same time that the
first cyclist will arive. At what point along the path
should he do that?
1 answer
1 answer