CBSE Class 9 Answered
Graph plotting...and numerical
Asked by rajat951 | 30 Jul, 2009, 08:15: PM
Expert Answer
Since 3 km/hr doesn't seem reasonable speed for second car, we assume a it's 30 km/hr and proceed.
The acceleration of the first car = (v – u)/t = (-52 km/hr)/ (5/3600 hr) = -37449 km/hr2
= -(52 km/hr)2/(2 x -37440 km/hr2) = 0.03611 km = 36.11 m
Similarly we can calculate for the second car, and we find that,
Stopping distance for second car = 41.6 m and second car goes further.
Regards,
Team,
TopperLearning.
Answered by | 08 Aug, 2009, 07:50: PM
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