Monday

November 24, 2014

November 24, 2014

Posted by **Matt** on Monday, November 5, 2012 at 3:56pm.

A. what is the negative acceleration?

B. How long after the brakes are first applied will it take to come to a complete stop?

C. How far does it travel before coming to a stop from when the brakes are first applied?

- Calculus -
**Damon**, Monday, November 5, 2012 at 4:11pmchange in velocity/change in time = (60-132)/3 = -24 ft/s^2

change in velocity = a t

0 - 132 = -24 t

t = 5.5 seconds to stop

d = Vi t + (1/2) a t^2

= 132(5.5) -.5(24)(5.5)^2

= 426 - 363 = 363 feet

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