Question

A car moving with a speed of 40 km/h can be stopped by applying brakes at least after 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance?
(a) 8 m
(b) 6 m
(c) 4 m
(d) 2.6 m

Answer 1

We can use the equation of motion to determine the minimum stopping distance because the equations of motion relate the speed of the object with distance, acceleration, and time. Here information about time is not given, so we will use the third equation of motion.

Complete step by step answer:
Two conditions are given in the question: In the first condition, the car's speed is $40{\rm{ km/h}}$, and after applying brakes, the car can be stopped after 2 m, and in the second condition, the speed of the car is $80{\rm{ km/h}}$.
Here the only speed of the car and distance is given, and time is not given to us. So, we will use the third equation of motion to calculate the required minimum stopping distance.
We will write the third equation of motion so that we can determine the acceleration of the car.
${v^2} = {u^2} + 2as$
Here, $v$ is the final speed of the car, $u$ is the initial speed of the car, $a$ is the acceleration of the ca and $s$ is the distance travelled by the car.
We know that the final speed of the car will become zero. Substitute v = 0 km/h and the values of the first condition in the above equation, therefore, we get,

0 = {\left( {40\;{\rm{km/h}} \times \dfrac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}} \times \dfrac{{1\;{\rm{h}}}}{{3600\;{\rm{sec}}}}} \right)^2} - 2a\left( {2\;{\rm{m}}} \right)\\\ a = \dfrac{{123.45\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}}}{{4\;{\rm{m}}}}\\\ a = 30.86\;{\rm{m/}}{{\rm{s}}^2} \end{array}$$ Now we obtained the value of the car's acceleration, so we will use the acceleration values with the given values of the second condition to get the minimum stopping distance when the speed of the car is 80 km/h. Therefore, from the third equation of motion, we get, $$\begin{array}{l} 0 = {\left( {80\;{\rm{km/h}} \times \dfrac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}} \times \dfrac{{1\;{\rm{h}}}}{{3600\;{\rm{sec}}}}} \right)^2} - 2\left( {30.86\;{\rm{m/}}{{\rm{s}}^2}} \right)s\\\ s = \dfrac{{493.83\;{{\rm{m}}^2}/{s^2}}}{{61.72\;{\rm{m/}}{{\rm{s}}^2}}}\\\ a = 8\;{\rm{m}} \end{array}$$ Therefore, the minimum stopping distance is 8 m, and option (a) is the correct answer. **Note:** In the question, the speed of the car is given in km/h, but the option is available in the meter, so do not forget to convert the units during the calculation. Otherwise, you will obtain a different answer without unit conversion, and it may vary from the given options.