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**Hint:**Issac Newton derived three equations of motion which describe the motion of a uniformly accelerated object. The equations depict the relations between the displacement, initial velocity, final velocity, time taken and acceleration of the object. Since the given car is mentioned to start its journey from rest, its initial velocity will be zero.

**Formulas used:**

The acceleration of a body is given by, $a = \dfrac{{v - u}}{{\Delta t}}$ where $v$ is the final velocity of the body, $u$ is its initial velocity and $\Delta t$ is the time interval.

The average velocity of a body is given by, ${v_{avg}} = \dfrac{{u + v}}{2}$ where $v$ is the final velocity of the body and $u$ is its initial velocity.

The distance travelled by a body is given by, $s = {v_{avg}} \times t$ where ${v_{avg}}$ is the average velocity of the body and $t$ is the time taken to cover the distance.

**Complete step by step answer:**

Step 1: List the three equations of motion derived by Newton.

The three equations of motion of a uniformly accelerated object as derived by Newton are:

$s = ut + \dfrac{1}{2}a{t^2}$

$v = u + at$

${v^2} - {u^2} = 2as$

In the above three equations, $s$ is the displacement of the object, $u$ is its initial velocity, $v$ is its final velocity, $a$ is its acceleration and $t$ is the time taken.

Step 2: Express the relation for the acceleration of the given car.

The above-mentioned car starts from rest and so its initial velocity is $u = 0{\text{m}}{{\text{s}}^{ - 1}}$ .

Its final velocity is given to be $v = 72{\text{km}}{{\text{h}}^{ - 1}}$ and the time interval is given to be $\Delta t = 10{\text{s}}$ .

Then the acceleration of the car can be expressed as $a = \dfrac{{v - u}}{{\Delta t}}$ -------- (1)

Substituting for $u = 0{\text{m}}{{\text{s}}^{ - 1}}$, $v = 20{\text{m}}{{\text{s}}^{ - 1}}$ and $\Delta t = 10{\text{s}}$ in equation (1) we get, $a = \dfrac{{20 - 0}}{{10}} = 2{\text{m}}{{\text{s}}^{ - 2}}$

The acceleration of the car is obtained to be $a = 2{\text{m}}{{\text{s}}^{ - 2}}$ .

Step 3: Express the relation for the average velocity of the car.

The average velocity of the given car can be expressed as ${v_{avg}} = \dfrac{{u + v}}{2}$ ---------- (2)

Substituting for $u = 0{\text{m}}{{\text{s}}^{ - 1}}$ and $v = 20{\text{m}}{{\text{s}}^{ - 1}}$ in equation (2) we get, ${v_{avg}} = \dfrac{{0 + 20}}{2} = 10{\text{m}}{{\text{s}}^{ - 1}}$

The average velocity of the car is obtained to be ${v_{avg}} = 10{\text{m}}{{\text{s}}^{ - 1}}$ .

Step 4: Express the relation for the distance covered by the car.

The distance travelled by the given car can be expressed as $s = {v_{avg}} \times \Delta t$ -------- (3)

Substituting for ${v_{avg}} = 10{\text{m}}{{\text{s}}^{ - 1}}$ and $\Delta t = 10{\text{s}}$ in equation (3) we get, $s = 10 \times 10 = 100{\text{m}}$.

**The distance travelled by the given car in the given time is obtained to be $s = 100{\text{m}}$.**

**Note:**While substituting values of physical quantities in an equation make sure that all the quantities are expressed in their respective S.I units. If this is not the case, then the necessary conversion of units must be done. Here, the final velocity of the car was expressed in the units of kilometres per hour as $v = 72{\text{km}}{{\text{h}}^{ - 1}}$. So we converted this into the units of metres per second, using the conversion relation $1{\text{km}}{{\text{h}}^{ - 1}} = \dfrac{5}{{18}}{\text{m}}{{\text{s}}^{ - 1}}$, as $v = 20{\text{m}}{{\text{s}}^{ - 1}}$ before substituting in equation (1).

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