5. Трамвай двигался прямолинейно равномерно со скоростью 54KM / 4 затем начал торможение с ускорением 0, 3M / (c ^ 2) Определите тормозной пусть и время торможения трамвая. Запишите уравнение скорости и постройте его график.
To find the braking distance and time of the tram, we can use the equation s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is time. Since the tram started braking from a speed of 54 km/h, we can convert this to m/s by dividing by 3.6, giving us an initial velocity of 15 m/s. The acceleration can be found by converting 0.3 m/(c^2) to m/s^2, which gives us an acceleration of approximately 8.33 x 10^-9 m/s^2. Now, we need to find the time it takes for the tram to stop, which we can find by setting the final velocity to 0 and solving for t. Plugging in the values, we get t = 15/(8.33 x 10^-9) = 1.8 x 10^9 seconds. This is approximately 57 years, which suggests that the tram will continue to decelerate for a very long time. As for the braking distance, we can plug in the values for t in the original equation, giving us s = (15)(1.8 x 10^9) + (1/2)(8.33 x 10^-9)(1.8 x 10^9)^2 = 24.3 billion meters. To put this into perspective, this is approximately three times the distance from Earth to Pluto! As a graph, the equation would look like a steep negative curve, with the distance increasing exponentially as time increases.
