You have to tackle these type of problem in steps: Solve for the first leg of the trip, and then solve for the second leg of the trip.
Use the following equation for the first leg of the trip:
Vf^2=Vi^2+2ad
Where
d=900m*(1/4)=225m
a=2.2 m/s^2
Vi=0m/s
and
Vf=?
Solve for Vf:
Vf^2=Vi^2+2ad
Vf^2=0+2(2.25m/s^2)*(225m)
Vf^2=1,012.5m^2/s^2
Vf=(1,012.5m^2/s^2)^1/2
Vf=31.8m/s^2
Vf for the first leg of the trip =Vi for the second leg of the trip. Solve for Vf of the second leg of the trip, which is the maximum speed of the trip.
Again, use the following equation for the first leg of the trip:
Vf^2=Vi^2+2ad
Where
d=900m*(3/4)=675m
a=0.750 m/s2
Vi=31.8m/s^2
and
Vf=?
Solve for Vf:
Vf^2=Vi^2+2ad
Vf^2=31.8m/s^2+2(0.750 m/s2)*(675m)
I'll let you do the rest.
A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at
rest (at x = 900 m). Through the first ¼ of that distance, its acceleration is +2.25 m/s2
. Through
the rest of that distance, its acceleration is – 0.750 m/s2
. Find its maximum speed?
1 answer
1 answer