Q70P - Fundamentals Of Physics

Question

Two particles move along an x axis. The position of particle 1 is given by $\mathbf{x}\mathbf{=}\mathbf{6}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{t}\mathbf{+}\mathbf{2}$ (in meters and seconds); the acceleration of particle 2 is given by $\mathbf{a}\mathbf{=}\mathbf{-}\mathbf{8}\mathbf{t}$ (in meters per second squared and seconds), and, at $\mathbf{t}\mathbf{=}\mathbf{0}$, its velocity is $\mathbf{20}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}$. When the velocities of the particles match, what is their velocity?

Step-by-Step Solution

Answer

At the time $1.05 s$the velocities are equal to$15.6 \mathrm{m}/\mathrm{s}$.

Step 1: Given data

The position of particle 1 is given by $x=6t^2+3t+2$.

The acceleration of particle 2 is given by$a=-8t$ .

Step 2: Relation between displacement, velocity, and acceleration

If you differentiate displacement and velocity equations, you can find the velocity and acceleration, respectively. Similarly, if you integrate acceleration and velocity equations, you can find velocity and displacement.

Step 3: Calculation for the time when their velocities are equal

The given equation is,

$X=6.00T^2+3.00t+2.00$

Differentiate the position of Particle 1 to find Velocity,

$$\begin{gathered} x=6.00t^2+3.00t+2.00 \\ {V}_1=\frac{dx}{dt} \\ =\frac{d\left(6.00t^2+3.00t+2.00\right)}{dt} \end{gathered}$$

Therefore,

${\mathrm{V}}_1=12.00\mathrm{t}+3.00\left(\mathrm{i}\right)$

It is given that,

$a=-80t$

Integrate the acceleration of particle 2 to find Velocity,

$$\begin{gathered} \int \mathrm{adt}=\int -8.00\mathrm{t} \\ =-\frac{8.00{\mathrm{t}}^2}{2}+\mathrm{C} \end{gathered}$$

(Note: C is the constant of integration)

$V_2=-4.00t^2+C$                                                                 (ii)

You have been given that, at $t=0,V_2=20m/s$

$\begin{array}{l}{V}_2=-4.00t^2+c\\ 20=-4.00\times 0^2+C\end{array}$

V₂can be rewritten as,

$V_2=-4.00t^2+C$

As to find, ‘t’ when their velocities are equal, equate $V_1=V_2$_. Therefore,

$$\begin{gathered} 12.t+3.00=-4.00t^2+20 \\ 4.00t^2+12.00t+3.00-20=0 \\ 4.00t^2+12.00t-17=0 \end{gathered}$$

Solving the quadratic equation, youcan get the two values of t, 

 

${\mathrm{t}}_1=1.05 \sec \mathrm{and} {\mathrm{t}}_2=-4.04$

Since time is always a positive quantity, take the positive value of the t. Therefore, at$1.05 \sec$ the values of velocities will be same.

Step 4: Calculation for the velocity

To find the Velocity at $t=1.05 \sec$, you put $t=1.05 \sec$ in either $V_1$or $V_2$, 

Now substitute the time value inequation (i).

$$\begin{gathered} V_1=12.00t+3.00 \\ =\left(12\times 1.05\right)+3.00 \\ {V}_1=V_2 \\ =15.6 m/s \end{gathered}$$

Hence, at $1.05 sec$both the velocities are same and has value as $15.6\mathrm{m}/\mathrm{s}$.