Help a couple questions I cant get!!

[fungulo]

1.To avoid a potential collision with wildlife crossing the road, the driver of a car travelling 105 Km/h fully applies the brakes to stop as quickly as possible. If the car stops 3.8 s after the brakes are applied, what is the braking distance?

 

2.A vehicle comes to a stop 10.0 s after the brakes are applied. While the brakes were applied, the vehicle travelled a distance of 75.0m. If the vehicles acceleration remained constant during the braking, what was the vehicles initial speed?

 

3.Based on a reaction time of 1.50 s and a braking rate of -5.85 m/s, the stopping distance of a vehicle travelling 90 Km/h would be?

 

4.A high performance car initially travelling 97.2 Km/h came to a stop in just 2.9 s. The mass of the car and its contents was 1850 kg. what was the magnitude of the average net force (braking force) on the car while it decelerated?

[Klaynos]

Are you familiar with the kinematic equations?

 

http://en.wikipedia.org/wiki/Equation_of_motion#Classic_version

[5614]

Use: [math]s = ut + \tfrac{1}{2} a t^2[/math]

s = distance travelled

u = initial speed

t = time

a = acceleration

 

and: [math]v^2 = u^2 + 2as[/math]

v = final speed

all other symbols are same as above.

 

For the reaction time one: if it takes you 1sec to react, and you're travelling at 10m/s, then you'll travel 10m before you start braking. So treat it as a normal question, but then add 10m on to account for your reaction time. I made these numbers up, work them out for your question.

 

4) presumably you have come across the equation: [math]F=\frac{d(mv)}{dt}[/math] in some form or another. F is the force, d(mv) is the change in momentum (also sometimes written as dp, as p=mv), and dt is the change in time. So for example if something of mass 1kg, at an initial speed of 2m/s, brakes to 0m/s, in 1sec, then: d(mv) = 1kg*2m/s - 1kg*0m/s = 2, and dt = 1sec, and then the constant force that is applied to obtain that constant deceleration (or momentum change, it's the same thing) is d(mv)/dt = 2/1 = 2N.

 

Clearly I haven't given you the answers, but I think I've given you all the tools and equations you need, so now it's up to you to at least make a start. If you get stuck then post where you're up to and we'll help you out from there.

[thedarkshade]

It's kinda weird again 5614 that our symbols differ. When we have to deal with such problems (which we did long time ago) we used [math]v_0[/math] for initial velocity and [math]v_1[/math] for final velocity. Now I have noticed that these symbols differ from one country to another, but isn't there a convention that regulates which symbol stands for what?

[swansont]

I assume part of it is the time one wishes to spend dealing with subscripts. Easy to write by hand, but more effort to format on a computer.

[5614]

... and sometimes I'd use v_f and v_i for final and initial velocity. For the equations I mentioned above we were supposed to be taught to use x instead of s, for distance, in our physics A Levels, but nearly all us students were used to s from maths mechanics, so we continued to use s.

 

I must say it is annoying having all this different notation for the same thing. In A Levels there were rarely minor differences between different subjects, but now I've come to university there's a lot of different symbols, and all us students, coming from different places and countries, are all used to different symbols from each other and the course!

 

You said E=U/l, I'd say E=V/d etc. Our quantum course uses [math]\psi(x,t)=a \cos(kx-\omega t + \phi)[/math], whilst our vibrations and wave course uses [math]\psi(x,t)=a \cos(\omega t -kx+ \phi)[/math]... how irritating!

[swansont]

And then you have the converse of the problem, with one symbol representing different things.

 

[math]\lambda[/math] , for example, can be wavelength, mean free path, decay constant, linear charge density, or eigenvalue (and maybe more)

[Klaynos]

I'm a firm believer in the symbols not mattering as long as you tell people what they are

 

As 5614 says I was taught vuats in maths and v_f and v_i in physics, but vuats got there first and I find vuats easier to remember and from there I know the 5 variables used in the equations, that I only need 3 of them to find the other 2 and can from there pull (with great effort) the equations from memory.

[fungulo]

thanks for those equations but i already tried them before and i could not get the awnsers. so i just need the awnseres please.

[Klaynos]

You wont get your answers here, we'll help but not provide answers.

 

Show us what you've got so far...

[thedarkshade]

Where's the problem fungulo?

Could you just post (LaTeX is possible) what you have been trying to do so we can, hopefully, check your work and tell you where were you wrong.

[5614]

Klay: it's suvat (sue-vat (rhymes with bat))! Not vuats!

 

fungulo: ok, so if you're totally stuck, here's a good way to start:

1) what do you want

2) what do you have

3) what can help you

4) check the units

5) do it!

 

So for question 1:

 

1) distance moved whilst under a constant deceleration (i.e. 's' in the above equations)

 

2) u=105km/h ; v=0km/h ; t=3.8s

 

3) v²=u²+2as -- nah, doesn't help, don't know 'a'

 

s=ut+½at² -- nah, doesn't help, don't know 'a'

 

ah, so this means I missed out an equation in my last post that I probably should have told you. That is: [math]a=\frac{v-u}{t}[/math], it's only for constant acceleration, but you can see where it comes from, it's saying the acceleration is the difference in speeds divided by the time it takes to do that.

 

So now you can use this to work out the acceleration, and then you can update (2) to include acceleration. Then you can use either of the two equations to calculate s. The acceleration will be negative, because it's deceleration aka negative acceleration.

 

4) note that the speed is in km/h. You could do the whole question in hours, although be careful because the time is in seconds. It would be easier, and a better, more common and the "official" way of doing it would be to convert that speed into m/s.

 

105km/h -(*1000 to account for the kilo)> 105,000m/h -(divide by 60 to go from hour to min> 1750m/min -(divide by 60 to go from min to sec)> [math]29\tfrac{1}{6}[/math]m/s.

 

5) now just stick all the numbers into the equations and type it in on a calculator (or do it in your head) and you'll get your answer.

 

Follow a similar process for the other questions. If you get stuck write down every formula you know, check this thread to see if you missed any, then do the "what to I want" and "what do I have" to see if you can use any of the equations to help you. Sometimes like in this question you'll need to use two formulae.