Question

A car starts from rest and travels for 5.2 s with a uniform acceleration of +1.9 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.6 m/s2. The breaks are applied for 1.90 s.

(b) How far has the car gone from its start?

This is the last question I need to do, and I'm like super stuck because I keep getting the answer wrong.

Answer 1

I understand that I do 2 separate problems and add them up together later, but I probably did something wrong. Perhaps plugging in the wrong numbers in?

Answer 2

what do you have?

Answer 3

Okay I used this formula :
x = v1(t) + 1/2(a)(t)^2

Answer 4

correct right?
Uhm, I used v1 = 0 (not sure about this)
& lpug the rest of the numbers in.

Answer 5

ok,

Answer 6

Is that correct?

Answer 7

number? \(x_1=\)

Answer 8

??

Answer 9

Oh the first x1 term I got 16.128

Answer 10

how? \(x_1 = \dfrac {1}{2}*(1.9)*(5.2)^2=25.688\)

Answer 11

oh opps. I was doing the practice problem. Haha my bad.

Answer 12

But yeah I got that answer from my previous one.

Answer 13

ok,next?

Answer 14

-1.52

Answer 15

opps

Answer 16

you have to find \(v_2\)

Answer 17

-2.8

Answer 18

v2? as velocity final?

Answer 19

Because for part (a) for the question it was asking for the velocity final and I got 6.84 m/s (which is correct)

Answer 20

you start at rest and get \(V_2\) then, from \(V_2\) you apply braking to stop.

Answer 21

it's not final, final is \(V_3 =0\) the time you have full stop

Answer 22

Ohh, I see what I did wrong. So for the second for 1.9s & -1.6 m/s^2, I use 6.84? or the v2 for the second part.

Answer 23

how can you get 6.84?

Answer 24

This is a from a previous question. Asking for the final velocity for total. So I got 6.84 m/s

Answer 25

(a) How fast is the car going at the end of the braking period?
-> answer 6.84 m/s

Answer 26

I use a different thing, I don't use v1 or v2 in my class. Instead I use Vf and VI

Answer 27

I don't. I apply \[V_2 = V_1 + at \] where \(V_1 =0 ~~at~~rest\)
\[V_2 = 1.9* 5.2 = 9.88 m/s

Answer 28

Okay, so I used that and plug for v1 for the second equation??

Answer 29

nope, for this part I apply \[V_3^2 = V_2^2 +2as \\s = \frac{V_3^2 -V_2^2}{2a}\]this a is -1.6

Answer 30

@ ^ @ the numbers are confusing. I have trouble organizing numbers. lol
Can you possibly do vf or vi? It's easier for me know where to go.

Answer 31

and I got \(x_2 = 30.50 m\)