Question

A driver needs to stop for a red light and slows down at a rate of 4m/s^2. it takes 100m to stop. (a) Find the time required to stop. (b) Find the initial velocity. (c) Plot x(t), v(t), and a(t). I need to be walk through this one.

Answer 1

Are you familiar with the kinematic equations of motion?

Answer 2

\(\sf v_f=v_i+a\Delta t\)

\(\sf x_f=x_i+v_o\Delta t +\dfrac{1}{2}a_x(\Delta t)^2\)

\(\sf v_f^2=v_i^2+2a_x\Delta x\)

Answer 3

I'm also not allowed to use any other equation except these:
\[v^{\rightarrow}=\frac{dr^{\rightarrow}}{dt}\]
\[a^{\rightarrow}=\frac{dv^{\rightarrow}}{dt}\]
\[v^{\rightarrow}=v _{0}^{\rightarrow}+at^{\rightarrow}\]
\[r^{\rightarrow}=r _{0}^{\rightarrow}+v _{0}^{\rightarrow}t+\frac{1}{2}a^{\rightarrow}t^{2}\]

Answer 4

ok so you used the \[r^{\rightarrow}\] one

Answer 5

okay, no problem. Use the last equation you posted to find the time.

Answer 6

wait, you dont know the initial velocity.

Answer 7

nope I have to figure that out too

Answer 8

so you can't use \(\sf v_f^2=v_i^2+2a_x\Delta x\)?

Answer 9

no =(

Answer 10

can we use the fact that the ending velocity is zero?

Answer 11

yeah, we need to use that fact, but without using that last equation i posted, all the equations have 2 unknown variables.

Answer 12

I guess we can use \(\sf v_f=v_i+a\Delta t\)

solve for \(\sf v_i\), and use that into \(\sf x_f=x_i+v_o\Delta t +\dfrac{1}{2}a_x(\Delta t)^2\)

Answer 13

yeah I see that. so I have a and I have the r?

Answer 14

then our only unknown is the time.

Answer 15

ok so for the first one will look like this
\[0=v^{\rightarrow}t+4t\]

Answer 16

theres no t with the initial velocity and depending which way you set up the axis, the acceleration will take a negative or positive sign. \(\sf 0=v_i-4\Delta t\)

Answer 17

my mistake

Answer 18

no worries, now just sub that into the other equation, and solve for t

Answer 19

ok let me do that real quick

Answer 20

okay, it should look like this \(\sf 100=4\Delta t^2 -\dfrac{1}{2}4(\Delta t)^2\)

Answer 21

what happened to the xi

Answer 22

well normally, you wanna set up your coordinate system so that the initial value is zero, so \(x_i=0\) and it's just omitted.

Answer 23

shouldn't it be more like this? \[(4\Delta t^{2})\Delta t\]

Answer 24

ok that makes sense

Answer 25

no it's fine the way it is

Answer 26

because \(v_i=4t\) substituing that in gives what i wrote earlier

Answer 27

no I get that but in the equation I have its vi*delta t

Answer 28

sorry about leaving suddenly, something came up and i had to go. Hope you got the question.