# Interpreting Velocity graphs

1. Oct 4, 2007

### kraaaaamos

1. The problem statement, all variables and given/known data

5. A car starts from xi = 10m at ti = 0s and moves with the velocity graph shown in figure on the right.
a. What is the object’s position at t = 2s, 3s, and 4s?
b. Does this car ever change direction? If so, at what time?

2. Relevant equations

V = d/t

3. The attempt at a solution

for t at 2 secs...
according to graph v= 4m/s

so position (d) = (t)(v)
= (2)(4)
= 8m?

am i doing it right?

OR

v = change d/change t
4m/s = (x-10)/(2-0)
4m/s = (x-10)/(2)
8m/s = x-10
x = 18m

is that correct?

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Last edited: Oct 4, 2007
2. Oct 5, 2007

### Astronuc

Staff Emeritus
Well this can't be right if the car starts at 10 m and adds distance. The slope of the plot of velocity vs time is negative, which indicates the car is decelerating, and the since the slope is constant, the deceleration is constant.

Certainly when the car has a negative velocity, it is reversing.

Is one familiar with integration?

if v(t) = d x(t)/dt, then

x(t) = $$\int_0^t\,v(t) dt\,+\,x(0)$$

This might be useful:

http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html