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July 30, 2014

July 30, 2014

Posted by **Praveen** on Tuesday, July 27, 2010 at 9:02am.

- 11th Grade Physics -
**bobpursley**, Tuesday, July 27, 2010 at 9:24ambreak the problem into three parts:

accelerating, constant speed, deaccelerating.

Let x be the accelerating distance, y be the constant speed distance, x/4 be the deaccelerating distance.

1000=x+y+x/4 or 5x/4 +y=1000, or y= 1000-5x/4

so solve for x first. 72km/hr = 20m/s

Vf^2=2ad

400=2*1*x or x=200 that leads to all the other distances.

time= sum of distances divided by avg velocity during that time.

time=200/10 +750/20 +50/10

check my work.

- 11th Grade Physics -
**Praveen**, Wednesday, July 28, 2010 at 2:58pmThe answer to this question is 62.5s. Can you please explain it properly what is that Vf^2 = 2ad

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