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July 25, 2014

July 25, 2014

Posted by **Shrawan** on Friday, July 15, 2011 at 10:07am.

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?

- Physics -
**Henry**, Friday, July 15, 2011 at 9:56pma. 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^2-Vo^2) / 2d,

a = (0-(16.67)^2 / 40 = 6.94m/s.

Ts = (vF-2Vo) / A,

tS = (0 - 33.33) / -6.94 = 4.80s.

b. Do you mean +-0.2% and +-0.1%?

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