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[1) 9.8 m12149 l2.25 A car moving with a speed of 40 km/hr can be stopped by applying breaks after atleast 2m. If the same car ismoving with a speed of 80 km/h. What is the minimurm stopping distanceCBSE-98)]12m814 m[1)8 m1314 m[416 m

Answer»

If F is retarding force and s the stopping distance then 1/2 m2 =Fs For same retarding force, s a v2 s_2/s_1 = (v_2/v_1 )^2 =((80 km/h)/(40 km/h))^2 =4 s2 = 4s1 = 4 × 2 = 8m



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