Car accelerates then brakes. What is its displacement?

[shizupple]

Homework Statement

A car starts from rest and travels for 5.5 s with a uniform acceleration of +1.4 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.9 m/s2. The breaks are applied for 1.70 s.
(a) How fast is the car going at the end of the braking period?
4.47 m/s (I got this part)
(b) How far has the car gone from its start?
? m

Homework Equations

1) V=Vi + at
2) x=Vit + 1/2at^2
3) V^2=Vi^2 + 2ax
4) a=V/t
5) v=x/t

The Attempt at a Solution

Vf^2=Vi^2+2a(x)
4.47^2=7.7^2+2(1.4)(x)
-14.04=X

I know this obviously can't be the answer. I'm just confused as to how to do this because there is two different accelerations. Do i find the avg acceleration by adding them up and dividing by two or do i find the displacement before braking then subtract the displacement after braking? I'm not sure which formula to use?

 

[Anden]

I would use equation 2 two times, and then adding the two results (for b)

 

[tiny-tim]

Welcome to PF!

Hi shizupple! Welcome to PF!

(try using the X² tag just above the Reply box )

(b) …
I'm just confused as to how to do this because there is two different accelerations. Do i find the avg acceleration by adding them up and dividing by two or do i find the displacement before braking then subtract the displacement after braking? I'm not sure which formula to use?

These constant acceleration equations only work for one acceleration at a time.

So use the first acceleration, to find a distance x₁.

Then use the second acceleration, to find a distance x₂.

The total distance will be x₁ + x₂.

 

[shizupple]

Hi shizupple! Welcome to PF!

(try using the X² tag just above the Reply box )These constant acceleration equations only work for one acceleration at a time.

So use the first acceleration, to find a distance x₁.

Then use the second acceleration, to find a distance x₂.

The total distance will be x₁ + x₂.

I actually tried this too. I did:
.5(1.4)(5.5)^2=21.175
.5(-1.9)(1.7)^2=-2.7455

I added them together without the - sign to get 23.9205... wasnt correct. So i tried subtracting the -2.7455 to get 18.4295 (which is not even logical) and wasnt correct.

 

[Anden]

The first part (t = 5,5 s) d = 0,5 a t^2, yields 21,175 as you stated.

But in the second you have a velocity Vi, you have forgotten to factor that into the equation

 

[shizupple]

Hmm I'm kind of confused on this. Yes there is initial velocity but you don't know how long there is initial velocity. It just says the car accelerated then broke. Which to me means it instantly broke so when you do the first part of the equation you would be multiplying (7.7)(0) which would be zero? I don't see how you would figure in the initial velicity of before it was started braking?

 

[Anden]

Hmm I'm kind of confused on this. Yes there is initial velocity but you don't know how long there is initial velocity. It just says the car accelerated then broke. Which to me means it instantly broke so when you do the first part of the equation you would be multiplying (7.7)(0) which would be zero? I don't see how you would figure in the initial velicity of before it was started braking?

t is the same in the different equation, 1,70 s. d denotes distance from the point of origin, and at the point of origin Vi = 7,7 so it's important that you factor it in.

Try to put Vi = 7,7 and t = 1,70 s along with a = - 1,9 m/s^2 into d = Vi t + 0,5at^2.

The variable t is the same for both Vi t and 0,5at^2, otherwise t would have to be separated from each other using indexes (like t1 and t2)

Also, you can see it like this: If the engine broke, the wheels would still be spinning for a while until the friction has stopped the vehicle. Just because the engine breaks, the car doesn't instantly stop.

I hope I'm not too confusing ;)

 

[shizupple]

Thats a good explanation. I didnt know you had to figure in the time to the initial velocity like that. Thanks for your help Anden! I got the answer (31.5195) on my last guess!