Question

for my problem a car is stopping at the stop light I know the velocity I know how long it takes it to stop. I solved where v=0 so the stopping point and another thing along the journey considering the position vs time is a parabola. I have two times because v=0 twice. I don't understand which one to use I have a large one and a small one.

Answer 1

Is there any equation?

Answer 2

\[\sf v_f=v_i+a\Delta t\]
\[\sf 100=(4\Delta t^2)\Delta t -\dfrac{1}{2}4(\Delta t)^2\]

Answer 3

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 4

I was solving for t and
t=7.071s (which is the answer) and t=24.5s
I'm just wondering why is it that one?

Answer 5

I am not sure how you solved it but here is my solution
v(final) = v(initial)+at
a = -4
let us assume it takes ta time for car to stop
v(final)=0
v(initial) = 4ta

Answer 6

Now distance = v(initial)t +1/2 at^2

Answer 7

if you solve you will get ta^2 =50

Answer 8

yeah that 7.071s

Answer 9

100 = 4ta^2 -2ta^2

Answer 10

Yes so only one positive value of t

Answer 11

humm. I didn't get the same. I got that number but I factored the t^2 out of an equation. I was left with 2t+1 too

Answer 12

Let me put all my work down so you can see and tell me where I messed up.

Answer 13

Ohh i see the mistake in your equation!

Answer 14

you messed up value of initial velocity

Answer 15

Oh I see it now! let me try again

Answer 16

vi=-4t
100=0+4t^2+4/2t^2
100=4t^2+2t^2
100=6t^2
I screwed up again somewhere...

Answer 17

it is 1/2at^2
and a =-4
so you should get 100=0+4t^2-4/2t^2

Answer 18

100=4t^2-2t^2
100=2t^2
50=t^2
7.1=t
thank you

Answer 19

finally :) np!