# Change in velocity

1. Oct 11, 2005

### NMW

2) the driver of a 600kg sports car, heading directly for a railroad crossing 250m away, applies the brakes i a panic stop. the car is moving at 40m/s, and the brakes can supply a friction force of 1200N. a) how fast is the car moving when it reaches the crossing?

- i tried finding the change in time for this, thinking i could then find displacment from that, but it ended up as 2.5hours!!
i then used f=ma and then average acceleration, got acceleration = 8.... or neg 8.... but then when i worked it out for change in velocity i got that the velocity was increasing, which cant be right seeing as the car is slowing down!!!???
any help appreciated!!

3) a metre-stick is found to balance at the 29.7cm mark when placed on a fulcrum, when a 50g mass is attached at the 10cm mark the fulcrum must be moved to the 39.2 cm mark for balance. what is the mass of the meter stick?

- just plain stuck on this, tried about 3 different ways of attempting the question, but couldnt find one that suited!! again, any help appreciated!!

:)

2. Oct 11, 2005

### Tom Mattson

Staff Emeritus
If you define the positive direction to be the direction of motion, then the acceleration is indeed negative. But it sure isn't -8. Just divide F by m and you get:

$$a=\frac{F}{m}=\frac{-1200N}{600kg}$$

You can finish that, right?

3. Oct 11, 2005

### hotvette

F = ma is the way to go. You just have to choose which direction is postive and keep everything consistent from there. Once you have F=ma you can integrate to get v and again to get x. This gives you three sets of equations specific to you problem that you can then use to get the answer.

4. Oct 11, 2005

### NMW

thanks

ok, i think i have got this one... i was just using the wrong value for force, which was throwing it all out.. once using -2m/s^2 i put that into the v^2=v(initial)^2 + 2a (x-x(initial) or something along those lines... it then came out as 24.5m/s, which fits with the rest of the calculations in the question!!
thanks!! :)