An airport tram carries passengers between two terminal buildings 500 m apart. The tram starts from rest and accelerations at 1.5 m/s^2 for 20 m. It then travels at a constant velocity for 450 m. Finally, it slows down for 30 m at an acceleration of 1 m/s^2 and comes to rest. How long does it take the tram to travel the entire 500 m trip between buildings?
1009
To find the time it takes for the tram to travel the entire 500 m trip between buildings, we need to break down the different phases of its motion and calculate the time for each phase separately.
Phase 1: Acceleration from rest for 20 m
The tram starts from rest and accelerates at 1.5 m/s^2 for 20 m. We can use the kinematic equation:
v = u + at
In this equation, v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.
For phase 1, the final velocity v is unknown, the initial velocity u is 0, the acceleration a is 1.5 m/s^2, and the distance traveled d is 20 m. We can rearrange the equation to solve for time:
v = u + at
v = 0 + 1.5t
t = v / 1.5
Substituting the values, we get:
t1 = 0.0225v
Phase 2: Constant velocity for 450 m
During this phase, the tram travels at a constant velocity, which means there is no acceleration. We can use another kinematic equation to find the time:
d = vt
In this equation, d is the distance traveled, v is the constant velocity, and t is the time.
For phase 2, the distance traveled d is 450 m, and the velocity v is unknown. We can rearrange the equation to solve for time:
t2 = d / v
Phase 3: Deceleration to rest for 30 m
The tram slows down and decelerates at 1 m/s^2 for 30 m. We can again use the kinematic equation:
v = u + at
For phase 3, the initial velocity u is unknown, the acceleration a is -1 m/s^2 (negative because it is deceleration), the distance traveled d is 30 m, and the final velocity v is 0 (because it comes to rest). Rearranging the equation to solve for time, we get:
v = u + at
0 = u + (-1)t
t = -u
t3 = -u
To find the total time, we add up the times for each phase:
total time = t1 + t2 + t3
To solve for the unknowns, let's consider the conditions of continuous motion. The initial velocity at the end of phase 1 is the same as the final velocity at the start of phase 2. Similarly, the initial velocity at the end of phase 2 is the same as the final velocity at the start of phase 3.
From phase 1:
t1 = 0.0225v
From phase 2:
t2 = d / v
From phase 3:
t3 = -u
Now, let's substitute these values into the total time equation:
total time = (0.0225v) + (d / v) + (-u)
Given that the initial and final velocities at each phase are the same, we can write:
u = v1 = v2
Since the tram starts from rest, v1 is 0, and since it comes to rest at the end, v2 is also 0. Let's substitute these values into the equation:
total time = 0.0225v + (d / v) + (-v)
Now we can solve for v. Since the distance traveled in phase 2 is 450 m, we can substitute this value into the equation:
total time = 0.0225v + (450 / v) + (-v)
To find the time, we need to minimize the total time equation by finding the value of v that yields the smallest time. Using calculus, we can take the derivative of the equation with respect to v and set it equal to zero:
d(total time) / dv = 0
Simplifying and solving this equation will give us the value of v that minimizes the total time. Then, we can substitute this value of v back into the total time equation to find the minimum time it takes for the tram to travel the entire 500 m trip between buildings.
Please note that the solution involves some complex mathematical calculations, and you may need to use numerical methods or a graphing calculator to find the value of v.