How Do You Calculate Position from a Velocity Graph?

[kraaaaamos]

Homework Statement

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?

Homework Equations

V = d/t

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?

 

[Astronuc]

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

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

am i doing it right?

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

 

[m2003]

I would first clarify the units for the velocity graph shown in the figure. Is it in meters per second (m/s) or kilometers per hour (km/h)? This information is important for accurately interpreting the graph.

Assuming the units are in m/s, here is my response:

a. According to the graph, at t = 2s, the car's position is at x = 18m. At t = 3s, the car's position is at x = 22m. At t = 4s, the car's position is at x = 26m.

b. Yes, the car does change direction. This can be seen on the graph where the velocity changes from positive to negative. This change in direction occurs at t = 3s.