Question

Рух двох матеріальних точок виражається рівнянням x1 = 20 + 2t – 4t2 та x2 = 2 - 2t + t2 (в системі СІ довжина вимірюється в метрах, час - в секундах). Знайти швидкості v1 і v2 в момент часу t = 5сек?

Answer 1

Ответ:

4*2=8

CI=t2

2-2t-t2=2-2t

t=5

5t*2t=10t

10t*2t=20t

Answer 2

Motion of Two Material Points

The motion of two material points can be described by the equations:

- x1 = 20 + 2t - 4t^2 - x2 = 2 - 2t + t^2

where x1 and x2 represent the positions of the two points at a given time t in the SI system (with length measured in meters and time in seconds).

To find the velocities v1 and v2 at t = 5 seconds, we need to differentiate the position equations with respect to time. The velocity is the derivative of position with respect to time.

Calculating the Velocities

Let's differentiate the position equations to find the velocities:

- v1 = dx1/dt - v2 = dx2/dt

Differentiating x1 and x2 with respect to t, we get:

- v1 = d(20 + 2t - 4t^2)/dt - v2 = d(2 - 2t + t^2)/dt

Simplifying these equations, we have:

- v1 = 2 - 8t - v2 = -2 + 2t

Now, we can substitute t = 5 seconds into these equations to find the velocities at that moment.

Substituting t = 5 seconds

Substituting t = 5 into the velocity equations, we get:

- v1 = 2 - 8(5) = 2 - 40 = -38 m/s - v2 = -2 + 2(5) = -2 + 10 = 8 m/s

Therefore, at t = 5 seconds, the velocity of point 1 (v1) is -38 m/s and the velocity of point 2 (v2) is 8 m/s.

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