A car is moving towards the check post with velocity 54 km/h. When the car is at 400 m from the check post, driver apples brake which is caused by deceleration of 0.3 m/s?
Find the distance of the car from the check post for 2 min after applying the brakes.

Hint: When brakes are applied in a car, the car begins to slow down and finally stop. Also when a car or any object slows down it means that the car is de-accelerating (which means acceleration is negative). Find out final velocity at t= 2 seconds and use it in the third equation of motion to find out the distance.

Complete step-by-step answer:
Given, Initial velocity of the car (u) = 54 km/h
$
   \Rightarrow \;\;54\,{\text{x }}\dfrac{5}{{18}} \\
   \Rightarrow \;\;15\,{\text{m/s}} \\
 $ ($\dfrac{5}{{18}}$ is a conversion factor from kilometer/hour to metre/sec)
Acceleration of the car = -0.3 ${\text{m}}{{\text{s}}^{ - 2}}$( car is de-accelerating)
T = 2 min = 2 x 60 = 120 seconds.
Final velocity after 120 seconds will be by using first law of motion we have,
$
  v = u + at \\
  {\text{or }}v = 15 - 0.3 \times 120 \\
   \Rightarrow \;v = 15 - 36 \\
   \Rightarrow \;v = - 21\,{\text{m/s}} \\
 $
Velocity cannot be negative as when a car accelerates, it stops. It cannot produce any negative velocity of its own.
Hence final velocity, v= 0.
Using the thing law of motion we have,
$
  {v^2} = {u^2} + 2aS \\
    \\
 $
Putting the values in above equation we have,
\[
   \Rightarrow {0^2} = {15^2} + 2 \times ( - 0.3) \times S \\
   \Rightarrow 225 = 0.6S \\
   \Rightarrow S = \dfrac{{225}}{{0.6}} = 375\,{\text{m}} \\
 \]
Distance of car from check post = 400 – 375 = 25m.

Hence, the answer is 25 metres.

Note: i) The question has asked distance away from pole and not distance travelled.
ii) After applying brakes, a car cannot move on its own and produce negative velocity. Hence v = 0.
iii) Deceleration means negative acceleration. Hence take signs of acceleration as negative.
iv) Always solve this type of question in SI units.