Posted by Praveen on Tuesday, July 27, 2010 at 9:02am.
break 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.
The answer to this question is 62.5s. Can you please explain it properly what is that Vf^2 = 2ad
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