Graph plotting...and numerical

Asked by rajat951 | 30th 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

The stopping distance = -u2/2a

= -(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  | 8th Aug, 2009, 07:50: PM

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