A car running at 72 kmh -1 is slowed down to 18 kmh-1 by the application of brakes over a distance of 20 m. Calculate i. retardation of car ii. total time in which car stops, and iii. total distance covered by it. Thank you so much

A car running at 72 kmh -1 is slowed down to 18 kmh-1 by the applica

1.	A car running at 72 kmh -1 is slowed down to 18 kmh-1 by the application of brakes over a distance of 20 m. Calculate i. retardation of car ii. total time in which car stops, and iii. total distance covered by it. Thank you so much
Answer» a step by step solution to your query:-Initial velocity u= 72 km/hr, final velocity v= 18 km/hr (convert km/hr into m/s by multiplying it by begin mathsize 12px STYLE 5 over 18 end style), this change in velocity OCCURS within the distance of s = 40 m.-As the velocity of the motion is decreasing, the acceleration is called retardation, which is numerically equal to the calculated acceleration 'a' but has a negative sign due to acceleration in opposite direction.-We get the value of retardation (in m/s2) on substituting the above VALUES in the formula,v2=u2+2as and write retardation in the negative terms of 'a'.-Further, on substituting the value of above calculated 'a', and the given values of 'u' and 'v' in the formula,v=u+at, (the value of 'a' will be in negative terms as it is retardation)we obtain the time 't' in SECONDS for which the brakes were applied by the given car.Hope this was helpful! Mark me brainiest