Find initial position via intergration???

An object moves in a straight line with an acceleration of 8m/s^2. If after 1 second it passes through point O and after 3 seconds it is 30 meters from O, find its initial position relative to O.

1 Answer
Jim H
May 7, 2018

Without knowing whether the position at 3 seconds is 30 meters left or right of O (or knowing whether the initial velocity is positive or negative) there are two correct answers.

Explanation:

#a = 8#

Using #v_0# for the initial velocity,

#v(t) = 8t+v_0#

Using #s_0# for the initial position.

#s(t) = 4t^2+v_0t+s_0#

We have been asked to find #s_0# as a function of O (the position at time #1# second.).

Using the values given for #t=1# and #t=3#, we get

#s(1) = 4 + v_0+s_0 = O# and

#s(3) = 36 +3v_0+s_0 = O +- 30#

Subtracting the first equation from the second yields:

#32+2v_0 = +-30# So

#2v_0 = -2# or #-62#

So #v = -1# or #v = -31#.

This gives us:
Solution 1

#s(t) = 4t^2-t+s_0# and, again using the information at #t = 1#, we find

#s(1) = 4-1+s_0 = O#,

so #s_0 = O-3#

Solution 2

#s(t) = 4t^2-3t+s_0# and, again using the information at #t = 1#, we find

#s(1) = 4-31+s_0 = O#,

so #s_0 = O-27#

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