Question

A car moves rectilinearly from station A to station B with an acceleration varying according to (a=b-cx) where b and c are positive constants and x is its distance from station A. find the distance between these stations if car stops at station B and it start from rest from A.

Answer 1

@LarsEighner

Answer 2

The only way I know to do this is by calculus. I'm pretty sure there are some ready-rolled formulas out there.

Let's see what it is telling us:

at time \(t_0= 0\), distance from the station A \(x_0=0\), and \(v=0\)

Answer 3

I'm trying to get my head around the acceleration formula.

\[a=b-cx\]

It says b and c are positive. So as x increases the acceleration decreases. Or in other words we have all the acceleration we are going to get to begin with (the max)

Answer 4

It also tells us that we stop a station B. so \(v_a = 0 \text { and } v_b = 0\).

Answer 5

ok, then

Answer 6

now calculus tells us velocity is the integral of acceleration

\[ \large \int b - cx dx = \int bdx - c\int x dx = bx - c({1 \over 2})x^2 + C \]

which might be looking something like a formula you know

v = bx + {1 \over 2}cx^2 + C

So what is C?

Answer 7

\[ \Large v=bx-{1\over 2}cx^2\]


(dammit, having latex problems.

We know velocity was zero to start with, we know it is zero at the end, but lookie here, it is expressed in displacement from A ( that is what x is). So you are not asked how long it takes to get there. You are not asked when you get there. You are asked for how far it is from A to B. Or in other words, what is x when the velocity is zero agian.

Low and behold, you have a second degree equation. It ought to have 2 roots (if there is any justice, although it could have fewer),

Answer 8

\[\large v = (x)(b-{1 \over 2}cx)\]

So that seems to be zero when x=0 (well we knew that, we started from zero velocity as zero displacement from A).

The other zero must tell us how far it is to station B.