Q73P - Fundamentals Of Physics

Question

At the instant the traffic light turns green, an automobile starts with a constant acceleration aof$\mathbf{2}\mathbf{.}\mathbf{2}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. At the same instant, a truck traveling with a constant speed of $\mathbf{9}\mathbf{.}\mathbf{5}\mathbf{m}\mathbf{/}\mathbf{s}$, overtakes and passes the automobile. (a)How far beyond the traffic signal will the automobile overtake the truck? (b)How fast will the automobile be traveling at that instant?

Step-by-Step Solution

Answer

(a) Automobile will overtake the truck $82m$beyond the traffic signal

(b) While overtaking the truck the velocity of the car is $19m/s$

Step 1: Given data

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

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

Step 2: Kinematic Equations of Motion

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.

Step 3:(a) Calculation of the point at which an automobile will overtake a truck from a traffic signal

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,

$$\begin{gathered} X_{covers by car}=X_{coverd by truck} \\ =V_{0}t+\frac{1}{2}a{t}^2 \\ =\frac{1}{2}a{t}^2 \end{gathered}$$

Therefore,

$X_{covered by car}=\frac{1}{2}a{\mathrm{t}}^2\left(\mathrm{i}\right)$

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

$X_{covered by truck}={\mathrm{v}}_0\mathrm{t}\left(\mathrm{ii}\right)$

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.

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

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

$$\begin{gathered} x_{coverd by car}=v_{0}t \\ =9.5\times 8.6 \\ =81.7 \\ \approx 82\mathrm{m} \end{gathered}$$

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

Step 4:(b) Calculation of velocity of the automobile, the moment it overtakes the truck

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

overtakes the truck,

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

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