A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platform is 5 m away from the door when the train starts to pull away and heads toward the door at acceleration of 1.2 m/s2. How far away from the door is the passenger after 3 seconds?
Help, I answered .4m and it was incorrect.
A) 2.7 m B) 5.4 m C) 0.40 m D) 2.3 m E) 0.90 m
the train has moved
distance=1/2 .6*3^2=2.7 m
the person has moved...
distance=1/2 *1.2*9=5.4 m OMG, the person has ran onto the tracks, into the train. Lets assume she was smart enough to not do that. I assumed she started at a distance perpendicular to the train. Maybe the author didn't mean that.
Lets try started at behind the door, next to the train.
train went 2.7m, person is now .4 m ahead of where the door was, so the distance is 2.3 m
sloppy wording on the problem, IMO
To solve this problem, we can use the equations of motion.
Let's break down the motion into two parts: when the train is accelerating uniformly and when the train is decelerating uniformly.
Step 1: Calculate the distance traveled during acceleration
In this phase, the passenger is not moving, and the train is accelerating at 0.6 m/s^2.
We can use the following equation:
d = ut + (1/2)at^2
Where:
- d is the distance traveled
- u is the initial velocity (0 m/s)
- a is the acceleration (0.6 m/s^2)
- t is the time (3 seconds)
Plugging in the values, we get:
d1 = 0(3) + (1/2)(0.6)(3^2)
d1 = 0 + (1/2)(0.6)(9)
d1 = 0 + 0.9
d1 = 0.9 meters
Step 2: Calculate the distance traveled during deceleration
In this phase, the train is decelerating at 1.2 m/s^2.
We can again use the equation:
d = ut + (1/2)at^2
Where:
- d is the distance traveled (we'll add it to d1)
- u is the initial velocity (0 m/s)
- a is the deceleration (-1.2 m/s^2 since it is in the opposite direction of acceleration)
- t is the time (3 seconds)
Plugging in the values, we get:
d2 = 0(3) + (1/2)(-1.2)(3^2)
d2 = 0 + (1/2)(-1.2)(9)
d2 = 0 + (-1.35)
d2 = -1.35 meters
Since we are interested in the distance from the starting point (ignoring direction), the total distance covered is the sum of d1 and absolute value of d2:
Total distance = d1 + |d2|
Total distance = 0.9 + |-1.35|
Total distance = 0.9 + 1.35
Total distance = 2.25 meters
Therefore, the passenger is 2.25 meters away from the door after 3 seconds. The correct answer is not provided in the options listed.
To find the distance traveled by the passenger after 3 seconds, we need to break down the problem into two parts: the initial acceleration of the train and the acceleration of the train after it starts moving.
First, let's calculate the distance traveled by the train during the initial 3 seconds:
Using the equation: distance = initial velocity * time + (1/2) * acceleration * time^2
- Initial velocity of the train (u1) = 0 m/s (starting from rest)
- Acceleration of the train (a1) = 0.6 m/s^2
- Time (t1) = 3 s
Plugging in these values, we get:
Distance traveled by the train during the initial 3 seconds (d1) = 0 m/s * 3 s + (1/2) * 0.6 m/s^2 * (3 s)^2
d1 = 0 + 0.9 m
Therefore, the train would have traveled 0.9 meters during the initial 3 seconds.
Next, let's calculate the distance traveled by the train during the final 3 seconds:
Using the same formula, but with different values:
- Initial velocity of the train (u2) = final velocity at the end of the initial 3 seconds (which can be calculated using v = u + at, where u = 0 m/s, a = 0.6 m/s^2, and t = 3 s)
=> u2 = 0 + 0.6 m/s^2 * 3 s = 1.8 m/s
- Acceleration of the train (a2) = 1.2 m/s^2 (since the train accelerates at a different rate after the initial 3 seconds)
- Time (t2) = 3 s (since we need to find the distance traveled during the final 3 seconds)
Plugging in these values, we get:
Distance traveled by the train during the final 3 seconds (d2) = 1.8 m/s * 3 s + (1/2) * 1.2 m/s^2 * (3 s)^2
d2 = 5.4 m
Therefore, the train would have traveled an additional 5.4 meters during the final 3 seconds.
To find the total distance traveled by the train, we add the distances traveled during the initial and final 3 seconds:
Total distance traveled by the train = d1 + d2 = 0.9 m + 5.4 m = 6.3 m
However, the question asks for the distance of the passenger from the door, not the total distance traveled by the train. Since the passenger is initially 5 meters away from the door, we subtract the total distance of the train from this initial distance:
Distance of the passenger from the door after 3 seconds = Initial distance - Total distance traveled by the train = 5 m - 6.3 m = -1.3 m
Since the distance cannot be negative, it means the passenger has already passed the door after 3 seconds. Therefore, the correct answer is none of the given options.