Question

Solving Ques. w/2 diff. formulas and getting 2 different ans... Frictional forces from brake pads are responsible for stopping an automobile. A car is moving at 35.0 m/s. If the car's brakes are capable of generating a deceleration rate of 6.00 m/s2, then how many meters will this vehicle move before coming to a stop? If I use the formula (Vf^2-Vi^2)/(2*acc) I get 102 meters but If I use the formula Acc=Velocity/Time I get 6=35/T with Time being 5.833. I then use the formula V=D/T and When I plug it in I get 35=D/5.833 with D being 204. Double of 102. Which ans is right and why?

Answer 1

For this response, I'll be using a couple symbols.
\(\Delta\) means "the change in", which is \((initial-final)\)
\(\bar\ \) goes over a variable to mean the average, like \(\bar a\) for average acceleration

I believe that your first approach is correct.

You have \(\Delta x=\dfrac{v_f^2-v_i^2}{2\bar a}\)
This formula comes from what I know to be
\(v_f^2=v_i^2+2\bar a\Delta x\), though I forget how it's derived at the moment. I want to look into that...



In the next one, you use the definition of acceleration, which can be expanded to
\(\bar a=\dfrac{v_f-v_i}{\Delta t}\quad\Rightarrow\quad\Delta t=\dfrac{v_f-v_i}{\bar a}=\dfrac{\Delta v}{\bar a}\)
The error comes in with the next assumption: \(\bar v=\dfrac {\Delta d}{\Delta t}\). This, as you must have seen, rearranges nicely to \(\Delta d=\bar v\ \Delta t\).
However, it cannot be used, because that equation is for an average velocity. And THAT'S the catch. You can't use the initial velocity where the formula is for the average.

Now, you might not have learned about how that \(v=\dfrac{\Delta d}{\Delta t}\) is the average velocity, \(\bar v\). That's probably because you learned about it when studying motion with \(constant\ acceleration\). Then, the velocity is always the same. Since it's always the same, it is always the average velocity.



If you were to use \(\dfrac{35-0}{2}=\dfrac D{5.833}\), where the left hand side is now \(\bar v\), you will get your 102.

Answer 2

. I then use the formula V=D/T.
That is the wrong. Car moves only with this velocity at the first instant then it's decreasing due to acceleration ,so the correct formula for the second method is:
you have t, Vo and a to calculate x:
\[x = V _{0}t- 0.5 a t ^{2}\].

Answer 3

and both will give you the same answer.