Question

A car is traveling at 33.0 m/s when the driver uses the brakes to slow down to 29.0 m/s in 2.50 seconds. How many meters will it travel during that time?

Answer 1

I think you use this kinematic equation. idk not totally sure
\[distance=v_i +.5at^2\]
your given time, initial velocity, and you can solve for acceleration given you have the change in velocity over a set time.

Answer 2

should work

Answer 3

we can derive it if you're not sure how the formula is figured out

Answer 4

Either way I wanna know.

Answer 5

what do you know at this point?

Answer 6

Well I know the givens. But I'm not sure which equation should I use and which would be easier. I don't know how to derive it either.

Answer 7

so what equations do you know ?

Answer 8

Well I'm using what Buger told me lol

Answer 9

smh, besides that?

Answer 10

None.

Answer 11

no clue about velocity nor acceleration?

Answer 12

Nope

Answer 13

LAUGHING OUT LOUD! for real?

Answer 14

lol yea hahha

Answer 15

try this vid https://www.youtube.com/watch?v=N6NPDoIoJWY
and we can expand later.

Answer 16

KK

Answer 17

it is a short and good summary

Answer 18

ya

Answer 19

ok soo....

Answer 20

watch it and try to learn

Answer 21

do some algebra

Answer 22

Velocity Initial is 33.0m/s right?

Answer 23

I got 82.5
Is that right??

Answer 24

correct about the initial velocity

Answer 25

Wht about the answer?

Answer 26

I didn't solve it

Answer 27

Oh.. Well do it?? Lol

Answer 28

Hi @YanaSidlinskiy !
How do you want me to help you?

Answer 29

Is tht answer right?

Answer 30

one min

Answer 31

Ok

Answer 32

Ok, the car should travel 77.5 meters

Answer 33

so 82.5 is wrong?

Answer 34

Yes it is.

Answer 35

There are two ways you can solve this question.

Answer 36

Will you show me how??

Answer 37

\(\sf Method~I\)
Use V=U+at to find the acceleration and then substitute this acceleration in any of the other equations of motion to find distance.

\(\sf Method~II\)
You must have noticed that the above method involves two steps and is often lengthy.
So I give you a trick equation
use \(\sf S= \frac{U+V}{2} \times t\)

You see no info about acceleration is required here

Answer 38

I did method 1

Answer 39

That's fine. Do it again. It must be a calculation error.

Answer 40

Ok well the time is obviously gonna be
2.50
IV= 33.0

Answer 41

Right?

Answer 42

correct

Answer 43

I don't get what would be the U though...

Answer 44

U = initial velocity, V = final velocity

Answer 45

Ohhh wow! Ok lol. So dumb. Ok and acceleration would be 29.0 because it slowed down which deaccelerated correct?

Answer 46

Nopes!!
Initial velocity is 33 m/s. Final velocity is 29 m/s because that's the velocity of the car after slowing down. You need to find out the acceleration using these data.

Answer 47

So??
29m/s= 33m/s(2.50m/s)?

Answer 48

How?

Answer 49

Lol idk. I'm sooooo confused rn

Answer 50

You know that,
Final velocity = Initial velocity + acceleration x time

Answer 51

Now tell me do you think you wrote correct equation?

Answer 52

No

Answer 53

Then rectify it

Answer 54

Idk what else to do to it

Answer 55

throwing in formulas do not help after all

Answer 56

Nope not at all.

Answer 57

It's simple yana!!
Just plug in the values dear.

29 = 33 + a x 2.5

Answer 58

Getting this?

Answer 59

Ok so acceleration is only when the rate or speed increases right?

Answer 60

Just be relaxed, everyone has to struggle initially. ;)
Kinematics is very easy..you'll see this gradually

Answer 61

Lol stop winking. Anywho... Ok can you walk me through this problem at least

Answer 62

can we do this gradually like from the simplest of the simplest

Answer 63

I feel like you failed to understand some crucial concepts prior to learning acceleration

Answer 64

"Acceleration" this term is used when speed increases and when speed decreases the term used is retardation.

Answer 65

But overall both are same physical quantity.

Answer 66

Ok.

Answer 67

decelerate and accelerate
not retardation

Answer 68

Ok smarty

Answer 69

I will create a post for you later

Answer 70

Thank you.

Answer 71

@nincompoop , both term are used

Answer 72

all this formula memorization is rubbish

Answer 73

Ok how would I continue to solve the problem?

Answer 74

Yes, find the unknown a from the above equation.

Answer 75

You really think I'm gonna read tht? Lol. And Abhi I don't know how!!!!!

Answer 76

It's a simple linear equation yana!
I know you know how to solve linear equations in one variables!!

Answer 77

yes, you should because that is part of what it will take to comprehend physical sciences concepts

Answer 78

29 = 33 + a x 2.5
Find a in the above equation.

Answer 79

what makes this different from your algebra lessons is that it tends to be more application and analytical than learning the mechanics of solving x or y

Answer 80

Idk

Answer 81

yana, how well did you do in your algebra lessons?

Answer 82

V = U + at


Where ,
V =final velocity = 29m/s
U = initial velocity = 33m/s
a= accelerate = ?
t = time = 2.5s

29 = 33 + a(2.5)
29 = 33 + 2.5a
29-33 = 2.5a
-4 = 2.5a
-4/2.5 = a
-1.6 m/s^2 =a

Since acceleration is negative we call this retardation.

Therefore , retardation = -1.6 m/s^2


Now, To find the displacement ,
\[s = ut + \frac{ 1 }{ 2 }at\]

Substitute the values and solve the equation and you'll get the distance traveled during this time.