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susus · Oct 21, 2010, 03:12 PM

An object is moving in a straight line with a constant acceleration. Its position is measured at three different times, as shown in the table below.
Time (s) | Position, (m)
44.00 | 8.900
46.00 | 17.900
48.00 | 38.500
Calculate the magnitude of the acceleration at t=46.00 s.

harum · Oct 21, 2010, 07:19 PM

If acceleration is constant, then why would it matter for what time point you calculate it?
Here is the equation for a body moving with a constant acceleration:
x(t) = x0 + v0*t + (1/2)*a*(t^2); here x(t) is position at a time t -- it is given, x0 is an initial position of the object -- it is unknown, v0 -- initial unknown speed of the object, and a -- unknown acceleration.

Plug the position and the time three times -- you will get three equations.

Three unknowns for three equations. Your system of three equations (one equation for each time point) will have a unique solution.

susus · Oct 22, 2010, 02:16 AM

I was not able to solve it :(

susus · Oct 22, 2010, 07:07 AM

I got it ! Thanks !

Unknown008 · Oct 22, 2010, 07:20 AM

Is this that difficult?

1. Time = 44 s, x = 8.9 m

We get:

$8.9 = x_o + u(44) + \frac12 a(44)^2$

$8.9 = x_o + 44u + 968a$

2. Time = 46 s, x = 17.9 m

$17.9 = x_o + u(46) + \frac12 a(46)^2$

$17.9 = x_o + 46u + 1058a$

3. Time = 48 s, x = 38.5 m

$38.5 = x_o + u(48) + \frac12 a(48)^2$

$38.5 = x_o + 48u + 1152a$

That's enough... let me resume the equations:

$8.9 = x_o + 44u + 968a$
$17.9 = x_o + 46u + 1058a$
$38.5 = x_o + 48u + 1152a$

Eliminate x_o first. Subtract equation 1 from 2 first, then 2 from 3.
1 from 2;
$9 = 2u + 90a$

2 from 3;
$20.6 = 2u + 94a$

From those two equations, we get:

$4a = 11.6$

$a = 2.9 ms^{-2}$

~~~~~~~~~~~~
Now, something to consider. I'll take a slightly different approach. We reset the times and distances. So, our object at time 0 is at position 0, with a new starting speed u.

First interval of time. Distance = 17.9-8.9 = 9 m. Time interval = 46-44 = 2 s.

$9 = u(2) + \frac12 a(2^2)$

$9 = 2u + 2a$

Second time interval and first combined. Distance = 38.5 - 8.9 = 29.6 m. Time interval = 48-44 = 4s

$29.6 = u(4) + \frac12 a(4^2)$

$29.6 = 4u + 8a$

Divide by 2;

$14.8 = 2u + 4a$

From those two equations, we get:

$2a = 14.8 - 9$

$a = 2.9 ms^{-2}$

Same answer, with less equations and a little more thought! :)

susus · Oct 22, 2010, 09:05 AM

Thanks
I solved it actually before !
I found easier way too . Than the two ways .
And no it's not difficult , but I have no physics background !

Unknown008 · Oct 22, 2010, 09:17 AM

Yes, there is yet another :p

Find twice the gradients of the three points.

Gradient of first two points = $\frac{17.9 - 8.9}{46-44} = 4.5$

Gradient of last two points = $\frac{38.5-17.9}{48-46} = 10.3$

Gradient between the two = $\frac{10.3 - 4.5}{2} = 2.9$

:p

And yes excon, it's THAT easy :)

susus · Oct 22, 2010, 09:18 AM

Yeah this one ;)