An automobile and train move together along
parallel paths at 20.3 m/s.
The automobile then undergoes a uniform
acceleration of −4 m/s2 because of a red light
and comes to rest. It remains at rest for 19.4 s,
then accelerates back to a speed of 20.3 m/s
at a rate of 1.95 m/s2.
How far behind the train is the automobile
when it reaches the speed of 20.3 m/s, as-suming that the train speed has remained at
20.3 m/s?
Answer in units of m
V = Vo + at
T1 = (V-Vo)/a = (0-20.3)/-4 = 5.075 s. to stop.
T2 = 19.4 s. @ rest.
T3 = (V-Vo)/a = (20.3-0)/1.95 = 10.4 s.
to regain speed.
Time Lost = T1+T2+T3 = 5.075+19.4+10.4 = 34.875 s.
d = 20.3m/s * 34.875s = 708 m. Behind
the train.
Well, cars and trains have a lot in common. They both have wheels, they both transport people, and they both occasionally have to stop for red lights. But when it comes to distances, they can be a little tricky. Now, I'm no mathematician, but let's see if I can put a smile on your face while we figure this out.
First, let's look at the part where the car is slowing down. We know it's accelerating at -4 m/s^2. Now, that's one speedy deceleration! It's like the car realized it forgot its keys or something.
Now, the car comes to a stop and takes a little break for 19.4 seconds. I hope it brought a good book or some snacks, because that's a longer break than most people get at work. But hey, we all need a breather sometimes.
After the break, the car starts accelerating again. This time, it's going at a rate of 1.95 m/s^2. It's like the car realized it had a deadline to meet and decided to get moving again. I can relate to that.
Now, let's answer the question. How far behind the train is the car when it reaches a speed of 20.3 m/s? Well, since they were moving at the same speed before the car hit that red light, we can assume they maintain the same distance throughout.
So, to find the distance, we need to figure out how long it takes for the car to reach a speed of 20.3 m/s again. We know the car's acceleration is 1.95 m/s^2, so we can use a little formula called v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time.
If the car's initial velocity is 0 m/s (because it came to a stop), and the final velocity is 20.3 m/s, we can rearrange the formula to solve for time. We get t = (v - u) / a.
Plugging in the values, we get t = (20.3 m/s - 0 m/s) / 1.95 m/s^2. That gives us a time of approximately 10.41 seconds.
Now, we need to find the distance the car travels in that time. We can use another formula: s = ut + (1/2)at^2, where s is the distance traveled.
Since the car started from rest (u = 0), we can simplify the formula to s = (1/2)at^2. Plugging in the values, we get s = (1/2) * 1.95 m/s^2 * (10.41 s)^2. That gives us a distance of approximately 107.36 meters.
So, when the car reaches a speed of 20.3 m/s, it's approximately 107.36 meters behind the train. I hope that clears things up, and remember, always wear your seatbelt, even if you're just doing math problems!
To find the distance between the automobile and the train when the automobile reaches a speed of 20.3 m/s, we need to calculate the total distance the train has traveled during the time the automobile was at rest.
Step 1: Calculate the time it took for the automobile to decelerate to rest.
Using the equation of motion v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance:
0² = (20.3)² + 2(-4)s
-410.09 = -8s
s = -410.09 / -8
s = 51.26125 m
The distance traveled by the automobile while decelerating to rest is 51.26125 m.
Step 2: Calculate the distance traveled by the automobile while it was at rest.
The distance traveled while at rest can be calculated using the formula distance = speed × time:
distance = 0 × 19.4
distance = 0 m
The distance traveled by the automobile while at rest is 0 m.
Step 3: Calculate the time it took for the automobile to accelerate back to a speed of 20.3 m/s.
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time:
20.3 = 0 + 1.95t
t = 20.3 / 1.95
t = 10.41026 s
It took 10.41026 seconds for the automobile to accelerate back to a speed of 20.3 m/s.
Step 4: Calculate the distance traveled by the train in that time.
The distance traveled by the train while the automobile was at rest can be calculated using the formula distance = speed × time:
distance = 20.3 × 10.41026
distance = 211.682478 m
The distance traveled by the train during the time the automobile was at rest is 211.682478 m.
Step 5: Calculate the total distance between the automobile and the train.
The total distance between the automobile and the train can be calculated by adding the distances found in steps 1, 2, and 4:
total distance = 51.26125 + 0 + 211.682478
total distance = 263.943728 m
Therefore, when the automobile reaches a speed of 20.3 m/s, it is approximately 263.943728 m behind the train.
To find the distance between the automobile and the train when the automobile reaches a speed of 20.3 m/s, we need to break down the problem into three parts:
1. Calculate the time it takes for the automobile to come to a stop.
2. Calculate the distance the automobile remains at rest.
3. Calculate the time it takes for the automobile to accelerate back to a speed of 20.3 m/s and the distance it travels during this time.
Let's start step by step:
1. Calculate the time it takes for the automobile to come to a stop:
The automobile has a uniform acceleration of -4 m/s^2. To find the time it takes to come to a stop, we can use the formula:
v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values:
0 m/s = 20.3 m/s + (-4 m/s^2) * t
Simplifying the equation:
-20.3 m/s = -4 m/s^2 * t
Solving for t:
t = -20.3 m/s / -4 m/s^2
t = 5.075 s
2. Calculate the distance the automobile remains at rest:
The automobile remains at rest for 19.4 s. Since it is not moving, the distance it remains at rest is 0 m.
3. Calculate the time it takes for the automobile to accelerate back to a speed of 20.3 m/s and the distance it travels during this time:
The automobile accelerates at a rate of 1.95 m/s^2 until it reaches a speed of 20.3 m/s. We can use the formula:
v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values:
20.3 m/s = 0 m/s + 1.95 m/s^2 * t
Simplifying the equation:
20.3 m/s = 1.95 m/s^2 * t
Solving for t:
t = 20.3 m/s / 1.95 m/s^2
t ≈ 10.41 s
Now, to find the distance traveled during this time, we can use the formula:
s = ut + 0.5at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values:
s = 0 m/s * 10.41 s + 0.5 * 1.95 m/s^2 * (10.41 s)^2
Simplifying the equation:
s = 0 + 0.5 * 1.95 m/s^2 * 108.3281 s^2
s ≈ 105.81 m
Adding up the distances:
Distance = 0 m + 0 m + 105.81 m
Distance ≈ 105.81 m
Therefore, when the automobile reaches a speed of 20.3 m/s, it is approximately 105.81 meters behind the train.