1. ## Application of Integration

Hi,
I am doing some home work and having problems with this. The teacher did not cover it well and the book that we are doing the homework from does not help at all.

The question is a car decelerates at 250 m/ss during an accident, the car was going 96 km/hr at impact, over what distance must an airbag stop a person in order to survive the crash.

so:
a= -250 t dt
v= 96km/hr X 3.6 = 345.6 m/s

a(t) = -250t dt
v(t) = -250t dt
vt = -125t^2 + C
345.6 = -125(0)^2 + C
c = 345.6

Do I have correct? I have no clue I know you go from accel to velocity then velocity to displacement.

Any help would be great, thanks.

2. ## Re: Application of Integration

It's easier to use Newton's Laws of motion (SUVAT) then it is to integrate.

You are given $u = 96 \times \dfrac{4}{9} \ ,\ v = 0 \ ,\ a = -250$ and want to find $s$. Thus you may use the equation $v^2=u^2+2as$ solving for s.

====================================

If you use calculus then $s = \int v\ dt \ ,\ v = \int a\ dt$

$v(t) = -250t dt$
You need to add a constant here since you integrated with respect to t: $v(t) = -250t + C_1$. You are told that at the time of collision (t=0) the speed is $v(0) = 96 \times \dfrac{4}{9} \text{ ms}^{-1}$

You can use this information to find $C_1$ from the known values of t and v(t)

To get displacement integrate again. $s(t) = -125t^2 + C_1t + C_2$

I'm not entirely sure how to find c_2.

3. ## Re: Application of Integration

Do I insert the 345.6 for T to find C then? I am a bit confused, but I do understand the equation but finding C is what I am not getting.
Thanks

4. ## Re: Application of Integration

The question is a car decelerates at 250 m/ss during an accident, the car was going 96 km/hr at impact, over what distance must an airbag stop a person in order to survive the crash.

so:
a= -250 t dt
v= 96km/hr X 3.6 = 345.6 m/s

both "a" and "v" are incorrect

a(t) = -250t dt
v(t) = -250t dt
vt = -125t^2 + C
345.6 = -125(0)^2 + C
c = 345.6

nope ...
first of all ...

96 km/hr = (96000 m)/(3600 sec) = 80/3 m/s

acceleration is a constant ...

a = dv/dt = -250

dv = -250 dt

v = -250t + C

at t = 0 , v = 80/3 m/s ...

v = (80/3) - 250t

the car comes to a stop when v = 0 at t = 8/75 sec

finally ...

$d = \int_0^{\frac{8}{75}} v(t) \, dt$

5. ## Re: Application of Integration

Ok worked it out and got it. I was stuck on getting time, but both of you helped out thanks.