91Ó°ÊÓ

Acceleration of an automobile is $2.2m/s^2$.

The speed of the truck is $9.5m/s$.

Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence motion. The equations which are used in the study are known as kinematic equations of motion.The kinematic equations relate the displacement, velocity, acceleration, and time with the motion of the objects.

The moment at which, the automobile overtakes the truck, the distance covered by the truck and the automobile will be the same.

So,it can be written that,

$$\begin{gathered} X_{coversbycar}=X_{coverdbytruck} \\ =V_{0}t+\frac{1}{2}a{t}^2 \\ =\frac{1}{2}a{t}^2 \end{gathered}$$

Therefore,

role="math" localid="1656302231139" $X_{coveredbycar}=\frac{1}{2}a{\mathrm{t}}^2\left(\mathrm{i}\right)$

As the car starts from rest i.e.$v_0=0$Use the same kinematic equation to write the distance traveled by the car.

role="math" localid="1656302348535" $X_{coveredbytruck}={\mathrm{v}}_0\mathrm{t}\left(\mathrm{ii}\right)$

Now, equate equations (i) and (ii).

$$\begin{gathered} v_{0}t=\frac{1}{2}a{t}^2 \\ {v}_0=\frac{1}{2}at \\ t=\frac{2v_0}{a} \end{gathered}$$

Substitute the value of the initial velocity of the truck and acceleration of the car.

$$\begin{gathered} t=\frac{2×9.5}{2.2} \\ =8.6\sec \end{gathered}$$

Substitute the value of time in equation (ii) to calculate the distance traveled.

$$\begin{gathered} x_{coverdbycar}=v_{0}t \\ =9.5×8.6 \\ =81.7 \\ ≈82\mathrm{m} \end{gathered}$$

So, the at$82\mathrm{m}$, automobile will overtake the truck as it passes the traffic signal.

Now, use the value of acceleration and time to calculate the speed of the car when it

overtakes the truck,

$$\begin{gathered} V_f=V_i+at \\ =2.2×8.6 \\ =18.92 \\ ≈19\mathrm{m}/\mathrm{s} \end{gathered}$$

The moment an automobile overtakes atruck, it has a velocity of magnitude$19\mathrm{m}/\mathrm{s}$.