acceleration question?

Question

A red car moving at 44.7 mph passes a blue car moving at 30.7 mph. (Use 1609 meters = 1 mile when converting to meters/second.) Seeing a stopped car in his lane up ahead, the driver of the red car applies the brakes just as he passes the blue car which causes the red car to slow down to a stop by losing 4.36 m/s every second.

(a) How much time (in seconds) passes before the two cars are side-by-side once again?
s




(b) When the cars are finally side-by-side, how do their average speeds compare for the time you calculated above? Please explain fully.

Answer 1

(a) The cars meet again when red's average velocity = blue's velocity.
v0 = vred(initial)
vred(ave) = 0.5*(v0+(v0+at)) = vblue
t = 2(vblue-v0)/a
I'll leave the conversions and arithmetic to you. (Remember 'a' is negative.)
(b) From the above their average speeds are equal; they covered equal distances in equal times.
EDIT: johnny solved for the time the velocities, not the average velocities, are equal. It takes twice as long for the average velocities and the displacements to become equal.

Answer 2

Blue car: (30.7mi/hr) x (1609m/mi) x (1hr/60min) x (1min/60sec) = 13.72 m/s
Red car: (44.7mi/hr) ... = 20 m/s, Accel = -4.36m/s

Use: v = v(0) + at
13.72 = 20 + (-4.36)t
-6.28 = -4.36t
t = 1.44s => Time when they are side by side.

b) S(avg) = total dist/ change in time

Blue: vel x time = dist
(13.72)(1.44) = 19.76m
S(avg,blue) = 19.76/1.44 = 13.72m/s

Red:
d = v(0)t + 1/2at^2
d = (20)(1.44) + 1/2(-4.36)(1.44)^2
d = 24.28m
S(avg,red) = 24.28/1.44 = 16.86m/s