A train starts from rest and accelerates uniformly, until it has traveled 3 km and acquired a velocity of 20 m/s The train then moves at a constant velocity of 20 for 400 s. The train then slows down uniformly at 0.065 m/s2 until it is brought to a halt. Calculate the total distance the train travels. Make a distance versus time graph and a speed versus time graph of the trains motion
If it takes 3 km to accelerate to 20 m/s, it will take the same distance (3 km) and time (300 s) to decelerate to zero. At constant speed, it travels 20 m/s*400s = 8000 m = 8 km. The total distance traveled is 8 + 3 + 3 = 14 km.
You will have to do your own graphing.
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Let's break down the problem into several steps to calculate the total distance the train travels:
Step 1: Calculate the time taken to reach a velocity of 20 m/s.
Using the equation v = u + at, where:
v = final velocity (20 m/s)
u = initial velocity (0 m/s)
a = acceleration (unknown)
t = time taken (unknown)
Rearranging the equation, we have:
t = (v - u) / a
Since the train starts from rest (u = 0), the equation simplifies to:
t = v / a
Plugging in the values, we get:
t = 20 / a
Step 2: Calculate the acceleration of the train.
Using the equation v^2 = u^2 + 2as, where:
v = final velocity (20 m/s)
u = initial velocity (0 m/s)
a = acceleration (unknown)
s = distance traveled (3 km = 3000 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Plugging in the values, we get:
a = (20^2 - 0^2) / (2 * 3000)
a = 400 / 6000
a = 0.0667 m/s^2 (rounded to 4 decimal places)
Step 3: Calculate the time taken to decelerate to a halt.
Using the equation v = u + at, where:
v = final velocity (0 m/s)
u = initial velocity (20 m/s)
a = deceleration (-0.065 m/s^2)
t = time taken (unknown)
Rearranging the equation, we have:
t = (v - u) / a
t = (0 - 20) / (-0.065)
t = 308.92 s (rounded to 2 decimal places)
Step 4: Calculate the distance traveled during uniform velocity.
The train moves at a constant velocity of 20 m/s for 400 s.
Using the formula d = v * t:
d = 20 * 400
d = 8000 m
Step 5: Calculate the distance traveled during deceleration.
Using the equation v^2 = u^2 + 2as, where:
v = final velocity (0 m/s)
u = initial velocity (20 m/s)
a = deceleration (-0.065 m/s^2)
s = distance traveled during deceleration (unknown)
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
s = (0^2 - 20^2) / (2 * (-0.065))
s = -400 / (-0.13)
s = 3076.92 m (rounded to 2 decimal places)
Step 6: Calculate the total distance traveled.
The total distance traveled is the sum of the distances traveled during acceleration, uniform velocity, and deceleration.
Total distance = distance during acceleration + distance during uniform velocity + distance during deceleration
Total distance = 3000 m + 8000 m + 3076.92 m
Total distance = 14076.92 m (rounded to 2 decimal places)
Now, let's create the distance versus time graph and the speed versus time graph:
Distance versus Time Graph:
- The graph will show three distinct parts:
1. The first part will be a curve representing the train's acceleration from rest to a velocity of 20 m/s.
2. The second part will be a straight horizontal line representing the constant velocity of 20 m/s.
3. The third part will be a curve representing the train's deceleration from 20 m/s to rest, gradually reaching zero velocity.
Speed versus Time Graph:
- The graph will also have three distinct parts:
1. The first part will be a straight line representing the train's acceleration from rest to a velocity of 20 m/s.
2. The second part will be a straight horizontal line representing the constant velocity of 20 m/s.
3. The third part will be a straight line representing the train's deceleration from 20 m/s to rest.
Note: The graphs can be sketched using appropriate scales for time and distance.
To calculate the total distance the train travels, we need to break down the motion into different stages and calculate the distance traveled in each stage.
Stage 1: Acceleration from rest to a velocity of 20 m/s.
To find the distance traveled during this stage, we can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity (20 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the distance traveled.
Solving the equation for s, we get:
s = (v^2 - u^2) / (2a)
Plugging in the values, the distance traveled during this stage is:
s = (20^2 - 0^2) / (2a)
Given that the acceleration is uniform, we need to find it first. We can use the equation:
v = u + at
Rearranging the equation to solve for the acceleration, a:
a = (v - u) / t
Since the train starts from rest, the initial velocity u is 0 m/s. Given that the final velocity v is 20 m/s and the time taken t is not given, we need to find the time using the equation:
s = ut + (1/2)at^2
Since the train has covered a distance of 3 km (3000 m), we can plug in the values and solve for t:
3000 = 0 + (1/2)a t^2
Rearranging the equation, we get:
t^2 = (2 * 3000) / a
t = sqrt((2 * 3000) / a)
Now, plugging in the values for v, u, and t in the acceleration equation, we can find the acceleration a:
20 = 0 + a * sqrt((2 * 3000) / a)
20 = sqrt(6000 / a)
400 = 6000 / a^2
a^2 = 6000 / 400
a^2 = 15
a = sqrt(15)
Plugging this value of a in the equation for the distance s, we can calculate the distance traveled during stage 1:
s = (20^2 - 0^2) / (2 * sqrt(15))
Stage 2: Constant velocity of 20 m/s for 400 s.
The distance traveled during this stage can be calculated by using the formula:
s = v * t
where v is the velocity (20 m/s) and t is the time (400 s).
Stage 3: Deceleration until the train is brought to a halt.
Similar to stage 1, we can use the equation of motion to find the distance traveled during this stage:
v^2 = u^2 + 2as
where v is the final velocity (0 m/s), u is the initial velocity (20 m/s), a is the acceleration (-0.065 m/s^2), and s is the distance traveled.
Solving for s, we get:
s = (0^2 - 20^2) / (2 * -0.065)
Now, we can calculate the total distance traveled by summing the distances traveled during each stage:
Total distance = Distance in stage 1 + Distance in stage 2 + Distance in stage 3
Finally, we can plot a distance versus time graph and a speed versus time graph based on the calculated distances and the given time intervals for each stage.