Cars A and B travel in a straight line. The distance of A from the starting point is given as a function of time by ${\mathbf{x}}_{\mathbf{A}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\alpha t}\mathbf{+}{\mathbf{\beta t}}^{\mathbf{2}}$, with $\mathbf{\alpha }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$and $\mathbf{\beta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. The distance of B from the starting point is ${\mathbf{x}}_{\mathbf{B}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{\gamma t}}^{\mathbf{2}}\mathbf{-}{\mathbf{\delta t}}^{\mathbf{3}}$, with $\mathbf{\gamma }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{80}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$and $\mathbf{\delta }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{3}}$. (a) Which car is ahead just after the two cars leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from A to B neither increasing nor decreasing? (d) At what time(s) do A and B have the same acceleration?

The given data can be listed below as,

• The function that defines the distance of car A from the start point is,$x_A\left(t\right)=\alpha t+\beta t^2$
• The value of$\alpha$ is, 2.60 m/s
• The value of$\beta$ is, 1.20 m/s²
• The function that defines the distance of car B from the start point is,$x_B\left(t\right)=\gamma t^2-\delta t^3$
• The value of$\gamma$ is, 2.80 m/s
• The value of$\delta$ is,$0.20\mathrm{m}/{\mathrm{s}}^2$

The velocity of the car has two important factors that are magnitude and direction and hence it is the vector quantity. Time taken and the displacement are the depending factors of the velocity.

It is known that the distance with respect to time is known as velocity. So, velocity can be obtained by taking the derivative of the function of position (distance) with respect to time. It is expressed as follows,

$v=\frac{dx}{dt}$

Here, x is the position of the object and t is the time taken.

(a)

Write the expression for the velocity of car A.

$v_A=\frac{d{x}_A}{dt}$

Substitute the value of$x_A$ in the above expression and solve.

$$\begin{gathered} v_A=\frac{d\left(\alpha t+\beta t^2\right)}{dt} \\ =\alpha +2\beta t \end{gathered}$$ …(1)

Write the expression for the velocity of car B.

$v_B=\frac{d{x}_B}{dt}$

Substitute the value of$x_B$ in the above expression and solve.

$$\begin{gathered} v_B=\frac{d\left(\gamma t^2-\delta t^3\right)}{dt} \\ =2\gamma t-3\delta t^2 \end{gathered}$$ …(2)

At the starting point that is, when t = 0, find the value of velocity of car A.

Substitute 0 for t in equation (1).

$$\begin{gathered} V_A=\alpha +2\beta \left(0\right) \\ =\alpha +0 \\ =\alpha \end{gathered}$$

At the starting point that is, when t = 0, find the value of velocity of car B.

Substitute 0 for t in equation (2).

$$\begin{gathered} v_B=2\gamma \left(0\right)-3\delta (0{)}^2 \\ =0+0 \\ =0 \end{gathered}$$

Thus, it can be observed that the velocity of car A is$\alpha$ at the starting point and the velocity of car B is 0 at the same point. So, it can be said that car B is behind car A.

(b)

Equate the expression for the position of both the cars and solve.

$$\begin{gathered} x_A=x_B \\ \alpha t+\beta t^2=\gamma t^2-\delta t^3 \\ \delta t^2\text{।}\left(\beta \gamma \right)t\text{।}\alpha =0 \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} \left(0.20\mathrm{m}/{\mathrm{s}}^3\right)t^2+\left(1.20\mathrm{m}/{\mathrm{s}}^2-2.80\mathrm{m}/{\mathrm{s}}^2\right)t+2.60\mathrm{m}/\mathrm{s}=0 \\ \left(0.20\mathrm{m}/{\mathrm{s}}^3\right)t^2-\left(1.60\mathrm{m}/{\mathrm{s}}^2\right)t+2.60\mathrm{m}/\mathrm{s}=0 \\ t=2.26\mathrm{s},5.73\mathrm{s} \end{gathered}$$

Thus, the time when both the cars are at the same position is at 0 s, 2.26 s, and 5.78 s.

(c)

Equate the expression for the velocity of both the cars, that is equate equation (1) and equation (2) and solve.

$$\begin{gathered} v_A=v_B \\ \alpha +2\beta t=2\gamma t-3\delta t^2 \\ 3\delta t^2+2(\beta -\gamma )t+\alpha =0 \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} 3\left(0.20\mathrm{m}/{\mathrm{s}}^3\right)t^2+2\left(1.20\mathrm{m}/{\mathrm{s}}^2-2.80\mathrm{m}/{\mathrm{s}}^2\right)t+(2.60\mathrm{m}/\mathrm{s})=0 \\ 3\left(0.20\mathrm{m}/{\mathrm{s}}^3\right)t^2+\left(3.2\mathrm{m}/{\mathrm{s}}^2\right)t+(2.60\mathrm{m}/\mathrm{s})=0 \\ t=1\mathrm{s},4.3\mathrm{s} \end{gathered}$$

Thus, the time when the position from car A to car B is neither incrementing nor decrementing is at 1 s and 4.3 s.

(d)

Write the expression for the acceleration of car A.

$a_A=\frac{d{v}_A}{dt}$

Substitute the value of$v_A$ in the above expression and solve.

$$\begin{gathered} a_A=\frac{d\left(\alpha +2\beta t\right)}{dt} \\ =2\beta \end{gathered}$$ …(3)

Write the expression for the acceleration of car B.

$a_B=\frac{d{v}_B}{dt}$

Substitute the value of$v_B$ in the above expression and solve.

$$\begin{gathered} a_B=\frac{d\left(2\gamma t-3\delta t^2\right)}{dt} \\ =2\gamma -6\delta t \end{gathered}$$ …(4)

Equate the expression for the acceleration of both the cars, that is equate equation (3) and equation (4) and solve.

$$\begin{gathered} a_A=a_B \\ 2\beta =2\gamma -6\delta t \\ t=\frac{\gamma -\beta }{3\delta } \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} t=\frac{\left(2.80\mathrm{m}/{\mathrm{s}}^2\right)-\left(1.20\mathrm{m}/{\mathrm{s}}^2\right)}{3\left(0.20\mathrm{m}/{\mathrm{s}}^3\right)} \\ =2.66\mathrm{s} \end{gathered}$$

Thus, the time when the acceleration of car A to car B is same is at 2.66 s.