Let's use the kinematic equations of motion to solve this problem. We'll need to use different equations for the different parts of the tram's motion.
First, let's find the distance traveled during the acceleration phase. We can use the equation:
d = v_it + 0.5a*t^2
where d is the distance traveled, v_i is the initial velocity (0 m/s), a is the acceleration (0.750 m/s^2), and t is the time (12.0 s).
d = 0 + 0.50.750(12.0)^2
d = 54.0 m
Now let's find the distance traveled during the constant velocity phase. We know that the tram travels at a constant speed for 20.0 s, so:
d = v*t
where v is the constant velocity and t is the time (20.0 s).
To find v, we can use the fact that the velocity acquired during the acceleration phase is maintained during the constant velocity phase. We can use the equation:
v_f = v_i + a*t
where v_f is the final velocity (the velocity acquired at the end of the acceleration phase), v_i is the initial velocity (0 m/s), a is the acceleration (0.750 m/s^2), and t is the time (12.0 s).
v_f = 0 + 0.750*(12.0)
v_f= 9.00 m/s
Now we can use the equation for distance traveled during the constant velocity phase:
d = vt
d = 9.0020.0
d = 180.0 m
Finally, let's find the distance traveled during the deceleration phase. We can use the equation:
d = v_it + 0.5a*t^2
where d is the distance traveled, v_i is the initial velocity (9.00 m/s), a is the acceleration (the deceleration, which is negative), and t is the time it takes to come to a stop.
To find a, we can use the fact that the deceleration is uniform and that the tram comes to a stop over a distance of 27.0 m. We can use the equation:
d = 0.5*(v_f + v_i)*t
where v_f is the final velocity (0 m/s), v_i is the initial velocity (9.00 m/s), and d is the distance (27.0 m).
27.0 = 0.5*(0 + 9.00)*t
t = 6.00 s
Now we can use t to find a:
27.0 = 9.006.00 + 0.5a*(6.00)^2
a = -0.750 m/s^2
Finally, we can use the equation fordistance traveled during the deceleration phase:
d = v_it + 0.5a*t^2
where d is the distance traveled, v_i is the initial velocity (9.00 m/s), a is the acceleration (the deceleration, which is negative), and t is the time it takes to come to a stop (6.00 s).
d = 9.006.00 + 0.5(-0.750)*(6.00)^2
d = 27.0 m
So the total distance traveled by the tram is the sum of the distances traveled during the three phases:
total distance = 54.0 + 180.0 + 27.0
total distance = 261.0 m
To find the time taken for the tram to travel that distance, we can add up the times for the three phases:
total time = 12.0 + 20.0 + 6.00
total time = 38.0 s
Therefore, the distance between the stops is 261.0 meters, and the time taken for the tram to travel that distance is 38.0 seconds.