Constant acceleration with known velocity and position

[lunarskull]

One-dimensional motion with constant acceleration

a body moving with uniform acceleration has a velocity of 12.0cm/s when its x coordinate is 3.00 cm. If its x coordinate 2.00s later is -5.00cm, what is the magnitude of its acceleration?

i am guessing that you use the equation for velocity in form of displacement, but I am not sure at all

some help on just picking an equation to use for this problem and starting the problem for me wud be greatly appreciated. more posts coming because I am so totally lost in this chapter.

 

[Päällikkö]

For uniform acceleration:
$$x = x_0 + v_0t + \frac{1}{2}at^2$$ (you can get the equation by integrating)

 

[lunarskull]

im still not getting it

i knew about this equation, but i still dunt no what I am doing wrong. i plugged in: (-5cm)=(3cm)+(12cm/s)+(1/2)(a)(2s)

 

[Päällikkö]

$$x = x_0 + v_0t + \frac{1}{2}at^2$$
Solve for a:
$$a = 2\frac{x-x_0-v_0t}{t^2}$$
$x, x_0, v_0, t$ are given.

 

[lunarskull]

...

i got -10 cm/s^2

but in the back of the book it says 16.

i put x=-5 $$x_0$$=3 $$v_0$$=12 and t=2s

i believe I am getting $$v_0$$ wrong

 

[Päällikkö]

I get (without units):
$$a = 2\frac{-5-3-12*2}{2^2} = -16$$
Did you notice it's $v_0t$, meaning $v_0$ times $t$?

 

[lunarskull]

wow thx, i wrote my equation on my equation sheet wrong.