In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A streetcar named Desire travels between two stations \(0.60 \mathrm{km}\) apart. Leaving the first station, it accelerates for \(10.0 \mathrm{s}\) at $1.0 \mathrm{m} / \mathrm{s}^{2}$ and then travels at a constant speed until it is near the second station, when it brakes at \(2.0 \mathrm{m} / \mathrm{s}^{2}\) in order to stop at the station. How long did this trip take? [Hint: What's the average velocity?]

We are given the following information: - Total distance between the stations: \(0.60 km = 600 m\) - Acceleration time: \(10 s\) - Acceleration: \(1.0 m/s^2\) - Deceleration: \(-2.0 m/s^2\) (braking, so negative value)

To find the distance covered during acceleration, we can use the formula \(d = vt + \frac{1}{2}at^2\), where \(d\) is distance, \(v\) is initial velocity, \(t\) is time, and \(a\) is acceleration. Since initial velocity is 0, we just have to compute \(d = \frac{1}{2}at^2\). \(d = \frac{1}{2}(1.0 m/s^2)(10 s)^2\) \(d= 50 m\) The distance covered during acceleration is 50 m.

We can use a similar approach for calculating the distance covered during deceleration. We should first find the final velocity during acceleration, which will be the initial velocity during deceleration. For this, we'll use the formula \(v_f = v_i + at\), where \(v_f\) is the final velocity, \(v_i\) is the initial velocity (0 in this case), \(a\) is acceleration, and \(t\) is time. \(v_f = 0 + (1.0 m/s^2)(10 s)\) \(v_f = 10 m/s\) Next, we need to find the time it takes to decelerate to a stop. We can use the formula \(t = \frac{v_f - v_i}{a}\), with \(v_i = 10 m/s\), \(v_f = 0\) (since the streetcar comes to a stop), and \(a = -2.0 m/s^2\) (deceleration). \(t = \frac{0 - 10 m/s}{-2.0 m/s^2}\) \(t = 5 s\) Now, we can calculate the distance covered during deceleration using \(d = vt + \frac{1}{2}at^2\): \(d = (10 m/s)(5 s) + \frac{1}{2}(-2.0 m/s^2)(5 s)^2\) \(d = 50 m - 25 m\) \(d = 25 m\) The distance covered during deceleration is 25 m.

Now that we have the distance covered during acceleration and deceleration, we can find the distance traveled at a constant speed: Constant speed distance = Total distance - (acceleration distance + deceleration distance) Constant speed distance = 600 m - (50 m + 25 m) Constant speed distance = 600 m - 75 m Constant speed distance = 525 m

We have the constant speed of the streetcar, which is 10 m/s. To find the time taken to cover the 525 m at constant speed, we can use the formula \(t = \frac{d}{v}\): \(t = \frac{525 m}{10 m/s}\) \(t = 52.5 s\)

Finally, we can add the time taken during acceleration, deceleration, and constant speed to get the total time taken for the trip: Total time = Acceleration time + Constant speed time + Deceleration time Total time = 10 s + 52.5 s + 5 s Total time = 67.5 s The total time taken for the streetcar's trip between the two stations is 67.5 seconds.