Request a call back

A car running at a speed of 72 km/hr is slowed down to 18 km/hr over a distance of 40 m.Calculate :(1) the retardation produced by the brakes.(2)time for which brakes are applied?
Asked by nisha_vini29 | 03 Jul, 2017, 12:24: PM
Here is 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 ), 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.