Question

Автомобиль из состояния покоя разгоняется с ускорением 1 м/с2, затем движется со скоростью 54 км/ч, а затем тормозит с ускорением -3м/с2 до полной остановки. Каково пере­мещение автомобиля? Время равномерного движения равно 20 с.

Answer 1

V - скорость

V=V0 + at

t=V/a

54км/ч=15м/с

t1=15/1=15c

t2=20c

t3=15/3=5c

S1=1*15^2/2= 112.5m

S2=15*20=300m

S3=3*5^2/2=37.5m

S(общ)=112.5+300+37.5= 450m

The end 

Answer 2

Calculation of Displacement

To calculate the displacement of the car, we need to consider the different stages of its motion: acceleration, uniform motion, and deceleration.

1. Acceleration: - The car starts from rest and accelerates with an acceleration of 1 m/s^2. - We can use the equation of motion: Sx = V0x * t + (1/2) * ax * t^2, where Sx is the displacement, V0x is the initial velocity, ax is the acceleration, and t is the time. - Since the car starts from rest, the initial velocity V0x is 0. - Plugging in the values, we get: Sx = 0 * t + (1/2) * 1 * t^2.

2. Uniform Motion: - The car moves at a constant speed of 54 km/h, which needs to be converted to m/s. - We know that 1 km/h is equal to 0.2778 m/s. - Therefore, the speed of the car in m/s is: 54 km/h * 0.2778 m/s = 15 m/s. - The car maintains this speed for a time of 20 seconds. - The displacement during uniform motion can be calculated using the equation: Sx = Vx * t, where Vx is the constant velocity and t is the time. - Plugging in the values, we get: Sx = 15 m/s * 20 s.

3. Deceleration: - The car decelerates with an acceleration of -3 m/s^2 until it comes to a complete stop. - Similar to the acceleration phase, we can use the equation of motion: Sx = V0x * t + (1/2) * ax * t^2. - Since the car comes to a stop, the final velocity Vfx is 0. - Plugging in the values, we get: 0 = V0x * t + (1/2) * (-3) * t^2.

Now, let's calculate the displacement of the car.

1. Acceleration phase: - The equation for displacement during acceleration is: Sx = (1/2) * t^2. - Plugging in the value of t (20 seconds), we get: Sx = (1/2) * (20)^2.

2. Uniform motion phase: - The equation for displacement during uniform motion is: Sx = Vx * t. - Plugging in the value of Vx (15 m/s) and t (20 seconds), we get: Sx = 15 m/s * 20 s.

3. Deceleration phase: - The equation for displacement during deceleration is: 0 = V0x * t + (1/2) * (-3) * t^2. - Since the car comes to a stop, the final velocity Vfx is 0. - Rearranging the equation, we get: V0x * t = (1/2) * (-3) * t^2. - Dividing both sides by t, we get: V0x = (1/2) * (-3) * t. - Plugging in the value of t (20 seconds), we get: V0x = (1/2) * (-3) * 20.

To calculate the total displacement, we need to sum up the displacements from each phase:

Total Displacement = Displacement during Acceleration + Displacement during Uniform Motion + Displacement during Deceleration

Now, let's calculate the values and find the total displacement of the car.