In figure, a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to x axis. At time t=0, the red car is at${\mathit{x}}_{\mathbf{r}}\mathbf{=}\mathbf{0}$and green car is at ${\mathbf{x}}_{\mathbf{g}}\mathbf{=}\mathbf{220}\mathbf{}\mathbf{m}$. If the red car has constant velocity of 20 km/hr, the cars pass each other atx = 44.5 m, and if it has a constant velocity of40 km/hr, they pass each other atx = 76.6 m. What are (a) the initial velocity? (b) the constant acceleration of the green car?

Position of red car,$x_r=0$

Position of green car,${\mathrm{x}}_{\mathrm{g}}=220\mathrm{m}$

Velocity of red car in first case,${\mathrm{v}}_1=20\mathrm{km}/\mathrm{hr}$

Point at which both cars cross each other in first case,${\mathrm{x}}_1=44.5\mathrm{m}$

Velocity of red car in second case,${\mathrm{v}}_2=40\mathrm{km}/\mathrm{hr}$

Point at which both cars cross each other in second case,${\mathrm{x}}_2=76.6\mathrm{m}$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the second kinematic equation, equations for the displacement of the both the green and the red car can be written.

By solving these equations simultaneously, either velocity or acceleration of the green car can be found. And finally, using this value of velocity in any one of above equation, the acceleration of the green car can be computed.

Formula:

The displacement in kinematic equation is given by,

$\begin{array}{l}∆x=V_{0}t++\frac{1}{2}a{t}^2\\ t=\frac{∆x}{v}\end{array}$

Let distance between the two cars be $x_g=220$at t = 0

Let $v_1$be the 20 km / h = 5.56 m/s corresponding to the passing point of ${\mathrm{x}}_1=44.5\mathrm{m}$and

$v_2$be the 20 km/h = 5.56 m/s corresponding to passing point of ${\mathrm{x}}_2=76.6\mathrm{m}$of the red car.

$x_g-x_1=v_g{t}_1+\frac{1}{2}a{t}_1^2$

Where,

$$\begin{gathered} t_1=\frac{x_1}{v_1} \\ =\frac{44.5}{5.55} \\ =8.00s \end{gathered}$$

Substituting the values of in the above equation,

Where,

$$\begin{gathered} {\mathrm{t}}_2=\frac{{\mathrm{x}}_2}{{\mathrm{v}}_2} \\ =\frac{76.6}{11.11} \\ =6.89\mathrm{s} \end{gathered}$$

Substituting the value of $v_2,t_2\mathrm{and}d$in the above equation,

$143.4=6.89{\mathrm{V}}_{\mathrm{g}}+23.74\mathrm{a}$

Solving above equations simultaneously, we get

The initial velocity of the green car

role="math" $$\begin{gathered} v_g=-13.9\mathrm{m}/\mathrm{s} \\ =-50\mathrm{km}/\mathrm{h} \end{gathered}$$

Negative sign means velocity is opposite to the velocity of red car or along the direction.

Therefore, initial velocity of the car is .

Using $v_g$in above equation, we get acceleration of the car

$$\begin{gathered} {\mathrm{v}}_{\mathrm{g}}=-13.9\mathrm{m}/\mathrm{s} \\ =-50\mathrm{km}/\mathrm{h} \\ \mathrm{a}=-2.0\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The negative sign means it is along the -x direction.

Therefore, acceleration of the green car is $-2.0\mathrm{m}/{\mathrm{s}}^2$