Homework Help: Help with velocity vector and friction questions

1. Apr 14, 2004

wikidrox

I can't quite get an answer for these damn questions. It feels like there is information missing.

1. Show that the minimum stopping distance for a car travelling at speed v is equal to v^2 / 2*coefficient of friction*g, where the coefficient of static friction is between tires and the road and g is acceleration due to gravity.

2. A ball rolls off the top of a stairway with a horizontal velocity of magnitude 1.5m/s. The steps are 20 cm high and 20 cm wide. Which step will the ball hit first?

2. Apr 14, 2004

Chen

1. Use this formula:
vf2 = v02 + 2ax
The distance is x, v0 is v and a is the car's deceleration caused by friction.

2. Let's create an axes system of X and Y, with the origin placed in the ball's inital position (top of stairs). The ball is traveling in a parabola. By analyzing the graph I attached, you can see that as soon as the ball's vertical displacement is bigger than or equal to the horizontal displacement, the ball will hit a step. (This is because the width and height of each step is equal. If this weren't the case, things would have been a bit more complicated.) So let's write the expressions for each displacement:

Δx = vt
Δy = gt2/2

Now solve this for t:

Δx = Δy
vt = gt2/2

One of the solutions is t = 0, because when the ball is first thrown it is on a step! So that's correct. The second solution is t = 2v/g. Now go back to the Δx expression and find how much the ball has managed to travel before hitting the step. For example, if the ball traveled 0.46 meters it means that it hit the 3rd step (if we don't count the step the ball was thrown from).

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Last edited: Apr 14, 2004
3. Apr 15, 2004

wikidrox

4. Apr 15, 2004

Chen

You don't need values, this is a parametric question. Find a (using what you know about the friction force) and then find the expression for x.

5. Apr 15, 2004