A streetcar named Desire travels between two stations 0.59 km apart. Leaving the first station, it accelerates for 10.0 s at 1.0 m/s2 and then travels at a constant speed until it is near the second station, when it brakes at 2.0 m/s2 in order to stop at the station. How long did this trip take?
To find the time it took for the streetcar to complete the trip, we need to break down and analyze the different phases of its motion.
Phase 1: Acceleration
The streetcar accelerates for 10.0 s at a rate of 1.0 m/s^2. To determine the final velocity, we can use the equation:
vf = vi + at
Where:
vf = final velocity
vi = initial velocity
a = acceleration
t = time
Since the streetcar starts from rest, the initial velocity is 0 m/s. Plugging in the given values:
vf = 0 + (1.0 m/s^2)(10.0 s)
vf = 10.0 m/s
Phase 2: Constant Speed
The streetcar maintains a constant speed between the two stations. To find the time taken for this phase, we can divide the distance traveled by the constant speed:
Distance = 0.59 km = 590 m
Constant Speed = 10.0 m/s
Time = Distance / Speed
Time = 590 m / 10.0 m/s
Time = 59.0 s
Phase 3: Deceleration
The streetcar decelerates at a rate of 2.0 m/s^2 to come to a stop near the second station. Using the equation for final velocity:
vf = vi + at
The final velocity is 0 m/s (as it comes to a stop). Again, the initial velocity is the constant speed from Phase 2, which is 10.0 m/s. Solving for time:
0 = 10.0 m/s + (-2.0 m/s^2) * t
-10.0 m/s = -2.0 m/s^2 * t
t = -10.0 m/s / -2.0 m/s^2
t = 5.0 s
Total Time = Time for Phase 1 + Time for Phase 2 + Time for Phase 3
Total Time = 10.0 s + 59.0 s + 5.0 s
Total Time = 74.0 s
Therefore, the trip took 74.0 seconds.