An automobile and train move together along
parallel paths at 28.8 m/s.
The automobile then undergoes a uniform
acceleration of −4 m/s
2 because of a red light
and comes to rest. It remains at rest for 63.8 s,
then accelerates back to a speed of 28.8 m/s
at a rate of 1.58 m/s
2
.
How far behind the train is the automobile
when it reaches the speed of 28.8 m/s, assuming that the train speed has remained at
28.8 m/s?
Answer in units of m
s₁=v²/2a₁=28.8²/2•4=103.7 m.
s₁=a₁t₁²/2 =>
t₁=sqrt(2s₁/a₁) =sqrt(2•103.7/4) =7.2 s.
t₂=63.8 s.
s₃=v²/2a₃=28.8²/2•1.58=262.5 m
s₃=a₃t₃²/2 =>
t₃ =sqrt(2s₃/a₃) =sqrt(2•262.5/1.58) =18.2 s.
Automobile:
s=s₁+s₃=103.7+262.5=366.2
t= t₁+t₂+t₃=7.2+63.8+18.2 = 89 s.
Train:
S=v •t=28.8•89=25 63.
Δs=S-s =2563.2-366.2=2197 m
To find the distance between the automobile and the train when the automobile reaches a speed of 28.8 m/s, we will use the equations of motion.
First, let's calculate the time it takes for the automobile to accelerate from rest to 28.8 m/s using the formula:
v = u + at
where v is the final velocity (28.8 m/s), u is the initial velocity (0 m/s), a is the acceleration (1.58 m/s^2), and t is the time.
Rearranging the equation, we get:
t = (v - u) / a
t = (28.8 - 0) / 1.58
t = 18.24 seconds
The time it takes for the automobile to accelerate to 28.8 m/s is 18.24 seconds.
Now, let's calculate the distance traveled by the automobile during this time interval. We'll use the equation:
s = ut + 0.5at^2
where s is the distance, u is the initial velocity (0 m/s), a is the acceleration (1.58 m/s^2), and t is the time (18.24 seconds).
Plugging in the values, we get:
s = 0(18.24) + 0.5(1.58)(18.24^2)
s = 0 + 0.5(1.58)(332.6976)
s = 0.5(1.58)(332.6976)
s = 261.6912 meters
So, the distance traveled by the automobile during acceleration is 261.6912 meters.
Since the train and automobile were initially moving together at the same speed of 28.8 m/s, the automobile will still be behind the train by this distance when it reaches the speed of 28.8 m/s.
Therefore, the automobile will be 261.6912 meters behind the train when it reaches the speed of 28.8 m/s.