You were driving a 1080 kg, 5.4 m long, blue car at a velocity of 29 m/s [about
ID: 1690386 • Letter: Y
Question
You were driving a 1080 kg, 5.4 m long, blue car at a velocity of 29 m/s [about 65 mph] . A 1040 kg, 7.2 m long, red car was 78 m ahead of you.The red car also had a velocity of 29 m/s . As you were driving, your cell phone rang. It only took you 3.45 s to locate your phone. However,
at the beginning of that brief 3.45 s time interval, the red car quickly began slowing down at a rate of 2.7 m/s**2 . As soon as you looked up and saw the red
car's brake lights, you realized that you also needed to slow down, but it took you an additional 1.41 s reaction time to get your foot on the brake pedal.
Both cars had the same braking acceleration.
Unfortunately, you quickly crashed into the red car.
a) What was the distance between the two cars after the 3.45 s time?
m
b) What was the distance between the two cars after the 1.41 s time?
m
c) How long after you put your foot on the brake pedal did you crash into the red car?
s
d) How fast was your blue car moving when you hit the red car?
m/s
e) How fast was the red car moving when you hit it?
m/s
Explanation / Answer
The speed of the blue car vB = 29 m/s
The speed of the red car vR = 29 m/s
The distance between the cars d = 78 m
The deceleration of the red car aR = 2.7 m/s^2
a)
The distance travelled by the red car at accceleration -2.7 m/s^2 in 3.45 s is
s1 = ut + 0.5*a*t^2
s1 = 29 m/s * 3.45 s + 0.5* (-2.7 m/s^2) * (3.45 s)^2
s1 = 84 m
The distance travelled by the blue car in 3.45 s is s2 = 29 m/s * 3.45 s
= 100 m
The distance between the two cars after
d' = (78 m + 84 m) -100 m = 62 m
b)
The speed of the red car after 3.45 s, vR' = u + a*t
= 29 m/s + (-2.7 m/s^2) * (3.45 s)
= 19.685 m/s
The distance travelled by the red car in next 1.41 s is
s3 = ut + 0.5*a*t^2
s3 = 19.685 m/s * 1.41 s + 0.5* (-2.7 m/s^2) * (1.41 s)^2
s3 = 25 m
The distance travelled by the blue car in 1.41 s is
s4 = ut
s4 = 29 m/s * 1.41 s
s4 = 40.9 m
The distance between the two cars d'' = (62 m + 25 m) - 40.9 m
= 46.1 m
c)
The speed of the red car after 1.41 s, vR'' = u + a*t
= 19.685 m/s + (-2.7 m/s^2) * (1.41 s)
= 15.878 m/s
Let after time t the cars get crashed.
The distance travelled by the blue car in t s is
s5 = ut + 0.5*a*t^2
s5 = 29t + 4.9 t^2
The distance travelled by the red car in t s is
s6 = ut + 0.5*a*t^2
s6 = 15.878 t + 4.9 t^2
The cars to be crashed when,
(s6 + 46.1) - s5 = 0
15.878 t + 4.9 t^2 + 46.1 -29t - 4.9 t^2 = 0
t = 3.5 s
It will take 3.5 s time for the cars to crash after applyinng the brakes on blue car
d)
The speed of the blue car after 3.5 s
vB' = u + a*t
= 29 m/s + (-2.7 m/s^2) * (3.5 s)
= 19.55 m/s
e)
The speed of the red car after 3.5 s
vR''' = u + a*t
= 15.878 m/s + (-2.7 m/s^2) * (3.5 s)
= 6.43 m/s
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