# Kenetics problems no clue how to slove can someone help?

ok so im new at the whole college physics course and im having problems with this ?

"A cab driver picks up a customer and delivers her 2.00 km away, on a straight route. The driver accelerates to the speed limit and, on reaching it, begins to decelerate at once. The magnitude of the deceleration is three times the magnitude of the acceleration. Find the lengths on the acceleration and deceleration phases."

looks like it should be and easy ? but im drawing blanks can anyone help?
thanks
dan

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Can you think of 3 equations in 3 unknowns that correspond to this problem ? Such a system is solvable.

thats the prob

the question what right out of the book

the only quantity given was the 2 km i know that the initial velocity is 0 of course but that isn't of much help.

x=1/2(v0 +vf)t
x=v0t + 1/2at^2

You don't really have the luxury of dealing with time. Try using 2*a*d = vf2 - vi2, and remember that the sum of the distances in question is 2.00km for the third equation.

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well i have 1 more equation i guess but im not sure how to solve

vf^2 = v0^2 + 2ax

would the final velocity be 0? i guess i still am not sure how to solve could you use trial and error to figure the distance?

technically if you were to accelerate for .5 km and decel for 1.5 km it would give u ur 2km but ya... im not very good in the physics department

dparent said:
well i have 1 more equation i guess but im not sure how to solve

vf^2 = v0^2 + 2ax

would the final velocity be 0? i guess i still am not sure how to solve could you use trial and error to figure the distance?
You're not just solving one equation for one variable. You're finding three valid equations that together contain 3 unknowns as well as data from the problem. If x1 is the first distance and x2 is the second distance, then you should get the three simultaneous equations v2 = 2ax1, v2 = 6ax2, and x1 + x2 = 2.00 .

is it possible to solve with the 3 unknowns? I don't no I appreciate the time you spent trying to explain it to me.
thank you

You're only interested in x1 and x2. Try substituting like variables for their equivalents in other equations in order to get an equation that consists of only one variable (either x1 or x2), then solve it algebraically.
For example, since the v's are the same variable (the maximum velocity) in the equations above, we can write 2ax1 = 6ax2.

k so i can get x1 alone by deviding by 2a so x1 = 3x2 im looking for the 2 lengths still in km right of which the decel is 3 times the length of accel there for accel should be x1 + x2 = 2km .5 km? and decel should be 1.5km ? or am i just totally not getting what you are saying lol...?