A Sunway train starts from rest at a station and accelerates at a rate of 2.0 ms^-2 for 10 s. It runs for constant speed for 50 s and slows down at a rate of 3.5 ms^-2 until it stops at the next station.
Find the total distance covered
Draw the velocity vs time graph for the whole trip
draw the velocity-time graph. The area below the graph is the distance covered. (area=velocity x time)
To find the total distance covered by the train, we need to calculate the distances during each segment of its motion and then add them together.
Segment 1: Acceleration
In this segment, the train starts from rest and accelerates at a rate of 2.0 m/s^2 for 10 seconds. To find the distance covered during this segment, we can use the equation:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
For acceleration, we have:
initial velocity = 0 m/s (since the train starts from rest),
time = 10 s, and
acceleration = 2.0 m/s^2.
Plugging these values in, we find:
distance1 = (0 * 10) + (0.5 * 2.0 * 10^2) = 100 m
Segment 2: Constant Speed
In this segment, the train runs at a constant speed for 50 seconds. Since the speed is constant, the distance covered is equal to the speed multiplied by the time:
distance2 = speed * time
We don't have the speed mentioned in the problem, so we can't calculate the exact distance for this segment.
Segment 3: Deceleration
In this segment, the train slows down at a rate of 3.5 m/s^2 until it stops. Here, the problem doesn't mention the time it takes for the train to stop, so we can't calculate the distance for this segment either.
Now, let's draw the velocity vs. time graph for the whole trip based on the given information.
- Segment 1: The graph will start at the origin (0,0) and will show a straight line with a positive slope (indicating acceleration) up to the point (10, 20 m/s). The line will be a curve gradually getting steeper.
- Segment 2: The graph will be a straight horizontal line at a positive velocity (indicating constant speed) starting from the point (10, 20 m/s) and going up to (60, 20 m/s), assuming the constant speed is 20 m/s.
- Segment 3: The graph will show a straight line with a negative slope (indicating deceleration) starting at (60, 20 m/s) and ending at the point where velocity equals zero (indicating the train has come to a stop).
Therefore, we are unable to calculate the total distance covered or draw the complete velocity vs. time graph due to missing information.