Graph plotting...and numerical

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.