Answer

Verified

52.8k+ views

**Hint:**The third equation of motion provides the final velocity of an object under uniform acceleration given the initial velocity and the distance traveled. By using the third equation of motion formula, apply the given values to find the minimum distance. When acceleration is constant, Velocity is proportional to time and Displacement is proportional to the square of time and Displacement is proportional to the square of the velocity.

**Formula Used:**

We will be using the formula of the third equation of motion i.e.,${v^2} - {u^2} = 2as$.

**Complete step by step answer:**

Given: The speed of the car $50km{h^{ - 1}}$

When brakes are applied final velocity becomes zero.

So we apply the third equation of motion

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

Where $u$ is the initial speed of the car $50km{h^{ - 1}}$

$v$ is the final speed of the car that is zero

$a$ is the acceleration

\[s\] is the distance that is $6m$

We know that, Distance $ = $average velocity $ \times $Time

Therefore, for constant acceleration

Average velocity$ = $\[\dfrac{{{\text{final velocity}} \times {\text{initial velocity}}}}{2}\]

Average velocity $ = \dfrac{{(v + u)}}{2}$

Hence Distance $s = \dfrac{{({v^2} - {u^2})}}{{2a}}$

Applying the value in the third equation of motion,

${v^2} - 50 = 2a \times 6$

The final velocity of the car is $v = 0$

Substituting the velocity of the car $v = 0$, and we can calculate the acceleration a is given by

\[a = \dfrac{{ - 2500}}{{2 \times 6}}\]

$a = \dfrac{{ - 2500}}{{12}}$

To find the minimum stopping distance

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

${100^2} = 2 \times (\dfrac{{ - 2500}}{{12}})s$

$s = 10000 \times \dfrac{6}{{2500}}$

$s = 24m$

Therefore the distance s is calculated

**Hence the minimum stopping distance of the car is option (D), $24m$.**

**Notes:**Dynamics and Kinematics are the two main descriptions of motion. Equation of motion can be used to derive the components like acceleration, velocity, and time. The relation between the components such as displacement, velocity, acceleration, speed, time, and distance are called Equations of motion. The third equation of motion is also called Laws of constant acceleration.

Recently Updated Pages

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Which of the following amino acids is an essential class 12 chemistry JEE_Main

Which of the following is least basic A B C D class 12 chemistry JEE_Main

Out of the following hybrid orbitals the one which class 12 chemistry JEE_Main

Other Pages

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Vant Hoff factor when benzoic acid is dissolved in class 12 chemistry JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main