Finding the distance with an acclerometer and a timer

[voxed]

Homework Statement

An accelerometer measures the acceleration of an object every 0.050s.
The initial velocity is 0 m/s.
The initial distance is 0 m/s.

The accelerations at the times (in m/s^2):
a(0.000) = 5
a(0.050) = 7
a(0.100) = 10
a(0.150) = 15
a(0.200) = 15
a(0.250) = 15
a(0.300) = 5
a(0.350)= -10
a(0.400)= -10

If possible, estimate the distance traveled by the object from t=0s to t=0.400s.

Homework Equations

$$v_{ave} = \Delta x / \Delta t$$

$$a_{ave} = \Delta v / \Delta t$$

$$x = x_0 + v_0 t + (1/2) a t^2$$

$$v^2 = v_0^2 + 2 a \Delta x$$

The Attempt at a Solution

I tried to find the slopes of the a(t) graph by having v(.025)=(7-5)/(.05-0) and v(.075)=(10-7)/(.1-.05). Then I tried to find the slope of the v(t) graph by x(.05)=(v(.075)-v(.025))/(.075-.025)=400m which is, of course, extremely high for the times and accelerations.

Another approach I thought of was to add up all of the recorded accelerations, divide them by the number of accelerations found, and use this average acceleration in

$$x = x_0 + v_0 t + (1/2) a t^2$$

as

$$x = 0 + 0 + (1/2) (average acceleration) (total time)^2$$

Are these correct approaches? Are there any other ways to do this that are more accurate?

Thanks.

 

[willem2]

You have to calculate the speed after every step

calculate the average acceleration in the first time interval.
use v = v_0 + a t to get the speed at the end of the first time interval
calculate the average speed in the first time interval
use x = x_0 + v_avg * t to get the position at the end of the first time interval

then repeat these steps for the next 7 time intervals