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July 29, 2014

July 29, 2014

Posted by **Candace** on Friday, January 22, 2010 at 12:21am.

a) find times t1 & t2 (s)

b)find velocity (ft/s)

At the 15800 ft mark, the sled begins to accelerate at -24 ft/s2.

(c) What is the final position of the sled when it comes to rest?

ft

(d) What is the duration of the entire trip?

- Physics -
**drwls**, Friday, January 22, 2010 at 7:11amt1 + t2 = 90

15,800 ft = acceleration distance + coasting distance

(1/2)*42*t1^2 + (42*t1)*t2 = 15800

Solve those two equations in the two unknowns, t1 and t2. The substitution method should work.

(b) V = 42 * t1 (after acceleration)

Use the value of t1 derived in part (a) to compute the initial V when deceleration begins.

(c) Let t' = deceleration time

24 t' = V

Solve for t'

Additional distance travelled while decelerating = (V/2)*t'

Add that to 15,800 for final position.

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