Cars A and B move in the same direction in adjacent lanes. The position x of car A is given in Fig. 2-30, from time$\mathit{t}\mathbf{=}\mathbf{0}$ to time $\mathit{t}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$. The figure’s vertical scaling is set by
${\mathit{x}}_{\mathbf{s}}\mathbf{=}\mathbf{32}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$. At $\mathit{t}\mathbf{=}\mathbf{0}$, car B is at $\mathit{x}\mathbf{=}\mathbf{0}$ , with a velocity of $\mathbf{12}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$and a negative constant acceleration ${\mathit{a}}_{\mathbf{B}}$. (a) What must${\mathit{a}}_{\mathbf{B}}$ be such that the cars are (momentarily) side by side (momentarily at the same value of x) at $\mathit{t}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$? (b)For that value of ${\mathit{a}}_{\mathbf{B}}$_, how many times are the cars side by side? (c) Sketch the positionx of car B vs timet on Fig. 2-30.How many times will the cars be side by side if the magnitude of acceleration ${\mathit{a}}_{\mathbf{B}}$ is (d) more than and (e) less than answer to part a.?

Position x of car A as per figure 2.30.

$x_s=32.0\text{m}$

Velocity of car B at $x=0$is $v_B=12\text{m/s}$.

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the formula for second kinematic equation, the negative constant acceleration ${\mathit{a}}_{\mathbf{B}}$that the cars are side by side at$\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$can be found. Further, using the formula for second kinematic equation, the values ${\mathit{a}}_{\mathbf{B}}$ fortother than$\mathit{t}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{}\mathbf{s}$using discrimination $\sqrt{{\mathbf{10}}^{\mathbf{2}}}\mathbf{-}\mathbf{2}\left(-20\right)\left(a_B\right)\mathbf{=}\mathbf{0}$can be determined, so that it can be decided whether cars will side by side.

Formula:

The displacement in kinematic equation is given by, $x=v_{0}t+\frac{1}{2}a{t}^2$.

With $v_a=\frac{12}{6}m/s=2m/s$and with$x_A=20\text{m}$, car B will be at$x_B=28\text{m}$when$t=4s$.

So,

$\begin{array}{rcl}{x}_A& =& 20+2t\\ 28\text{m}& =& \left(12\frac{\text{m}}{\text{s}}\right)\left(4\text{s}\right)+\frac{1}{2}{a}_B{\left(4\text{s}\right)}^2\\ a_B& =& -2.5{\text{m/s}}^{\text{2}}\end{array}$

Therefore, a negative constant acceleration $a_B$ that the cars are side by side at $t=4.0\text{s}$.

To find the values of t other than $t=4\mathrm{s}$ using the value $a_B=-2.5{\text{m/s}}^{\text{2}}$condition is $x_A=X_B$.

$\begin{array}{rcl}20+2t& =& 12t+1/2a_B{t}^2\\ 1.25t^2-10t+20& =& 0\end{array}$

There are two roots to this quadratic equation, unless discriminate $\sqrt{{10}^2-4\left(20\right)\left(1.25\right)}=0$since this discriminate is zero cars are side by side only once at $t=4s$.

Below is the graph showing the displacement of the both cars against the time.

If $a_B\ge 2.5$then there will be no real solutions to${10}^2-4\left(20\right)\left(a_B\right)<0$.

Therefore, the cars are never side by side.

If $\left(a_B\right)⩽2.5$then

We have,

So, it has two real roots.

Therefore, cars are side by side at two different times.