finding position, acceleration from a velocity graph?

Question

The velocity v(t) of some particle is plotted
as a function of time on the graph below.
The scale on the horizontal axis is 3 s per
division and on the vertical axis 8 m/s per
division.

graph: http://img.photobucket.com/albums/v321/Daesgesage/question.jpg

Initially, at t = 0 the particle is at x0 =
37 m.

What is the position x of the particle
at time t = 12 s? Answer in units of m.

What is the particle’s acceleration? Answer
in units of m/s^2.

Don't understand how to solve at all. Please give equation and explain. PLEASE. I have an hour till due. :-(

Answer 1

Initially (at time 0):

t = 0s

x = 37m

(from the graph) v = 32m/s



at t = 12s, (from the graph again, v = 8m/s)



The equation relating the two is:

V2^2 - V1^2 = 2xacceleration x (X2 - X0)



Where:

V2 = velocity at 12s (8m/s)

V1 = velocity at 0s (32m/s)

X2 = position at 12s (what we're looking for)

X1 = position at 0s (37m)

acceleration = acceleration. =)



So we need to find the acceleration first:

acceleration = (change in velocity)/(change in time)



Looking at the blue line, this is equal to the slope (rise)/(run) of the line.

the equation for slope is:

(Y2 - Y1)/(X2 - X1)



Where Y2, Y1 are the velocities at points 2 and 1,

X2, X1 are the times (in sec) at points 2 and 1.



choosing two points:

(X1, Y1) --> (0, 32)

(X2, Y2) --> (12, 8)

substitute:

(8 - 32) / (12 - 0) = -24 / 12 = -2m/s^2



The acceleration is negative because the particle is slowing down. (Recall that it started at 37m/s but became 12m/s after 12 secs)



Ok. We have the acceleration, we can now substitute it in our equation:

V2^2 - V1^2 = 2xacceleration x (X2 - X0)



V2 = velocity at 12s (8m/s)

V1 = velocity at 0s (32m/s)

X2 = position at 12s (what we're looking for)

X1 = position at 0s (37m)

acceleration = -2m/s^2



(8m/s)^2 - (32m/s)^2 = 2 x (-2m/s^2) x (X2 - 37m)

(64 - 1024)m^2/s^2 = (-4m/s^2) x (X2 - 37)m

-960m = (X2 - 37)m

X2 = -923 m



(Please check the values as I don't have a calculator right now, but the equations should be valid.)

Answer 2

ok lets try this. if y=mx+b

---> (y-b)/x=m

b=8

y=32

x=12

m=2



y=(2)x+8

Since we know that velocity is the first derivative of position, then that must mean that position is the first integrand of velocity.



y=f(x)



integrating f(x)=2x+8 (dx)



eq. A) x^2 +8x



at 12 sec. x=12 from the graph

soooo plug in 12 into eq. A) (2)(x=12)^2 +(8x=96)



288+96=384 meters



Buttt, you also said that it started at 37 meters away from somewhere, so just add 37 + 384 =421 meters





Ok, to find acceleration just take the derivative of the same

f(x)=2x+8



which is 2 and because of the slope, you are decelerating 2 m/s^2

Answer 3

gradient of postion time graph is velocity time graph gardient of velocity time graoh is acceleration time graph