Physics
posted by Shrawan .
An automobile travelling with a speed of 60km/hr, can apply brake to stop within a distance
of 20m. If the car is going twice as fast then calculate the stopping distance and stopping time?
(b) Given R1 = 5.0 ± 0.2 and R2 = 10.0 ± 0.1. calculate the total resistance in parallel with possible
% error?

a. The stopping distance is proportional to the square of the initial velocity:
Ds = 2^2 * 20m = 80m = Stopping distance.
The stopping time is proportional to the initial velocity:
Vo = 60000m/h * (1h/3600s) = 16.67m/s.
a = (Vf^2Vo^2) / 2d,
a = (0(16.67)^2 / 40 = 6.94m/s.
Ts = (vF2Vo) / A,
tS = (0  33.33) / 6.94 = 4.80s.
b. Do you mean +0.2% and +0.1%? 
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