two buddies going to sac town. On a bright sunny day , two friends are on a road trip and drive in two different cars. The two vehicle, Car A and Car B, move along interstate 80 in two different lanes going toward Sacramento.
1.) Car b is 200 m ahead of Car A and is traveling with a constant velocity of 30.0 m/s.
2.) Car A has veloctity 40.0 m/s. The driver observes Car B and applies her brakes, causing a constant acceleration of -0.100 m/s^2. In other words, Car A starts to slow down.
A.) what is the time when Car A passes Car B?
b) what is the time when Car B passes Car A?
c.) what is the velocity of Car A as a function of time?
can u please show which equations u used? plese help
cattbarf
March 15, 2010 at 12:08 pm
You can envisage the two cars as going the x-direction. Car b is at x=200 and moving at 30m/sec. Car a is a x=0, has been moving at 40 m/sec, and starts a constant deaccleration at that point.
Velocity of car A= 40-0.1 t
Distance of car A = 40t- 0.05 t^2
Distance of car B = 30t + 200.
Set theset two equal to each other and solve. The smaller t will be where A passes B and the larger t will be when B passes A.
gummstein
March 15, 2010 at 12:40 pm
S(a)=ut+1/2(a)(t^2)
S(b) = 30t
at t=0, S(a)=0, S(b)=200
when car A passes car B, or car B passes car A, S(a)= S(b)
40t-0.05(t^2)=30t + 200
800t-t^2=600t+4000
t^2-200t+4000=0
The solution for t is answer to both parts
v=u+at
V(a)=30-0.1t