Two cars are moving evenly on a chasse towards each other. Modules of their speeds are equal to 36

| May 29, 2021 | education

Two cars are moving evenly on a chasse towards each other. Modules of their speeds are equal to 36 km / h and 20 m / s. Determine the speed of the first car relative to the second and the second relative to the first.

V1 = 36 km / h = 10 m / s.

V2 = 20 m / s.

V12 -?

V21 -?

According to the law of relativity of velocities, the speed of a body relative to a stationary frame of reference Vн is the vector sum of the speed of a moving frame of reference relative to a stationary Vпн and the speed of a body relative to a moving frame of reference Vп: Vн = Vпн + Vп.

The fixed frame of reference is the earth, the movable frame of reference is the second car.

Therefore, the speed of the second car relative to the ground V2 is the speed of the moving frame of reference Vнп: Vнп = V2.

The speed of the first car relative to the second V12 is the speed of the body, relative to the moving frame of reference Vп: Vп = V12.

The speed of the first car relative to the ground V1 is the speed of the body relative to the stationary frame of reference Vн: Vн = V1.

V1 = V2 + V12.

V12 = V1 – V2.

Let’s draw the coordinate axis in the direction of the first car, Since the second car is moving towards it, then V12 = V1 – (-V2) = V1 + V2.

V12 = 10 m / s + 20 m / s = 30 m / s.

The speed of the second car relative to the first V21 will also be the sum, only directed in the opposite direction of the axis: V21 = – 30 m / s.



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