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ICSE Class 10 Answered

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
answered-by-expert Expert Answer
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 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!
Answered by Abhijeet Mishra | 03 Jul, 2017, 03:02: PM
ICSE 10 - Physics
Asked by kadtej83 | 27 Jul, 2019, 06:18: PM
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