Question

Two particles move along an x axis. The position of particle 1 is given by x = 9.00t^2 + 6.00t + 5.00 (in meters and seconds); the acceleration of particle 2 is given by a = -8.00t (in meters per seconds squared and seconds) and, at t = 0, its velocity is 20.0 m/s. When the velocities of the particles match, what is their velocity?

Answer 1

Answer on Question #51374, Physics, Mechanics | Kinematics | Dynamics

Question

Two particles move along an x axis. The position of particle 1 is given by

$$x = 9.00t^2 + 6.00t + 5.00 \text{ (in meters and seconds)};$$

the acceleration of particle 2 is given by $a = -8.00t$ (in meters per seconds squared and seconds) and, at $t = 0$, its velocity is $20.0 \, \text{m/s}$. When the velocities of the particles match, what is their velocity?

Solution

Let's compute the velocity of the particle 1.

$$x_1(t) = 9t^2 + 6t + 5$$ $$V_1(t) = \left(x(t)\right)' = \left(9t^2 + 6t + 5\right)' = 18t + 6$$

Velocity of the particle 2 can be expressed as:

$$V_2(t) = \int a(t) \, dt, \text{ where } V_0 \text{ is the initial velocity of the particle 2}.$$

We know that:

$$a(t) = -8t$$ $$V_2(0) = 20, \text{ then:}$$ $$V_2(t) = \int -8t \, dt = -4t^2 + C$$ $$V_2(0) = C = 20$$ $$V_2(t) = 20 - 4t^2$$

So, we should solve the equation:

$$\begin{array}{l} V_{1}(t) = V_{2}(t) \\ 18t + 6 = 20 - 4t^{2} \\ 4t^{2} + 18t - 14 = 0 \\ 2t^{2} + 9t - 7 = 0 \\ D = 81 + 56 = 137 \\ t_{1} = \frac{-9 + \sqrt{137}}{4} t_{2} = \frac{-9 - \sqrt{137}}{4} \\ t_{2} < 0 \Rightarrow t_{1} \text{ is a unique solution} \\ t_{1} \approx 0,676(s) \\ \end{array}$$

**Answer**

$$0,676s$$

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