A commuter train travels between two downtown stations. Because the stations are only 1.36 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the travel time t between the two stations by accelerating at a1 = 0.100 m/s2 for a time t1 and then immediately braking with acceleration a2 = -0.410 m/s2 for a time t2. Find the minimum time of travel t and the time t1.
To find the minimum time of travel, we need to determine the values of t and t1 that minimize the total time for the journey.
Let's break down the problem into two parts:
1. Acceleration phase:
During the acceleration phase, the train is accelerating with an acceleration of a1 = 0.100 m/s^2. We need to find the time t1 for this phase.
Using the formula for distance traveled during constant acceleration:
d = v0*t + (1/2)*a*t^2
Where:
d = distance covered (1.36 km or 1360 m)
v0 = initial velocity (0 m/s, as the train starts from rest)
By substituting these values, we can rearrange the formula to find t1:
1360 = 0*t1 + (1/2)*0.100*t1^2
2720 = 0.050*t1^2
t1^2 = 2720/0.050
t1^2 = 54400
t1 = sqrt(54400)
t1 ≈ 233.24 seconds
So, the time spent during the acceleration phase is approximately 233.24 seconds.
2. Deceleration phase:
During the deceleration phase, the train is braking with an acceleration of a2 = -0.410 m/s^2. We need to find the time t2 for this phase.
Using the same formula as above, but this time with the negative acceleration, we get:
1360 = 0*t2 + (1/2)*(-0.410)*t2^2
1360 = (-0.205)*t2^2
t2^2 = 1360/(-0.205)
t2^2 ≈ -6649.76
But since time cannot be negative, this means that the deceleration phase is not needed. The train will reach the destination without applying the brakes.
Therefore, the minimum time of travel, t, is equal to t1, which is approximately 233.24 seconds.
To find the minimum time of travel and the time taken for acceleration (t1), we can apply the laws of motion.
Let's start by applying the equations of motion to the first part of the motion where the train is accelerating:
Using the equation of motion: d = v0 * t + 0.5 * a * t^2
where,
d = distance traveled,
v0 = initial velocity (in this case, 0 m/s),
a = acceleration,
t = time taken,
Substituting the given values in the equation, we get:
1.36 km = 0 * t1 + 0.5 * 0.1 * t1^2
Simplifying this equation, we get:
1.36 km = 0.05 * t1^2
Converting kilometers to meters, we have:
1.36 km = 1360 m
1360 m = 0.05 * t1^2
Dividing both sides by 0.05, we get:
27200 = t1^2
Taking the square root of both sides, we obtain:
t1 ≈ 164.97 seconds
Now, let's find the time taken for the train to decelerate (t2). Since the train must stop at the destination, we know that the final velocity is 0 m/s.
Using the equation of motion: v = v0 + a * t
where,
v = final velocity (0 m/s),
v0 = initial velocity,
a = acceleration,
t = time taken,
Substituting the given values in the equation, we get:
0 = 0.1 * t1 - 0.41 * t2
Simplifying this equation, we have:
0.1 * t1 = 0.41 * t2
Substituting the value of t1 we found earlier, we get:
0.1 * 164.97 = 0.41 * t2
16.497 = 0.41 * t2
Dividing both sides by 0.41, we obtain:
t2 ≈ 40.24 seconds
Now, let's find the total time of travel (t) by adding t1 and t2:
t = t1 + t2
≈ 164.97 seconds + 40.24 seconds
≈ 205.21 seconds
Therefore, the minimum time of travel is approximately 205.21 seconds and the time taken for acceleration (t1) is approximately 164.97 seconds.