Question

Два тела движутся согласно законам х₁=-4 + 2t +t² и х₂=6 - 8t + t². Чему ровна относительная скорость тела в момент встречи?

Answer 1

В задании не указано как движутся тела, буду считать, что они движутся навстречу друг другу.

Скорости тел - это производные от пути

V1 = x1' = 2 + 2t

V2 = x2' = -8 +2t

Момент встречи найдем из уравнения:

-4 + 2t +t² = 6 - 8t + t²

2t + 8t = 6 + 4

10t = 10

 t = 1 сек

V1(1) = 2+2*1 = 4

V2(1) = -8+2*1 = -6

Разные знаки говорят о направлении скоростей.

Относительная скорость  V = V1 - V2 = 4 - (-6) = 10 

Answer 2

Calculating Relative Velocity at the Moment of Encounter

To calculate the relative velocity of the two bodies at the moment of encounter, we can find the derivative of the position functions with respect to time and then evaluate it at the time of encounter.

The position functions for the two bodies are given as: - Body 1: \( x_1 = -4 + 2t + t^2 \) - Body 2: \( x_2 = 6 - 8t + t^2 \)

Finding the Derivatives of the Position Functions

Let's find the derivatives of the position functions with respect to time to obtain the velocities of the two bodies.

The derivative of \( x_1 \) with respect to time \( t \) is: \[ v_1 = \frac{dx_1}{dt} = 2 + 2t \]

The derivative of \( x_2 \) with respect to time \( t \) is: \[ v_2 = \frac{dx_2}{dt} = -8 + 2t \]

Calculating the Relative Velocity at the Moment of Encounter

To find the relative velocity at the moment of encounter, we need to find the time \( t \) when the two bodies meet. This occurs when their positions are equal, so we can set \( x_1 = x_2 \) and solve for \( t \).

\[ -4 + 2t + t^2 = 6 - 8t + t^2 \] \[ 2t + 4t = 6 + 4 \] \[ 6t = 10 \] \[ t = \frac{10}{6} = \frac{5}{3} \]

Now that we have the time of encounter, we can find the relative velocity by evaluating the difference in velocities at this time.

The velocity of body 1 at \( t = \frac{5}{3} \) is: \[ v_1 = 2 + 2\left(\frac{5}{3}\right) = \frac{16}{3} \]

The velocity of body 2 at \( t = \frac{5}{3} \) is: \[ v_2 = -8 + 2\left(\frac{5}{3}\right) = \frac{2}{3} \]

Relative Velocity at the Moment of Encounter

The relative velocity of the two bodies at the moment of encounter is the difference between their velocities at that time: \[ v_{\text{relative}} = v_2 - v_1 = \frac{2}{3} - \frac{16}{3} = -\frac{14}{3} \]

Therefore, the relative velocity of the two bodies at the moment of encounter is -14/3 units of distance per unit of time.