Based on the variables in your question, the following equation will be used.

d=vt+1/2at^{2}

where d is distance,

v is initial velocity,

t is time, and

a is acceleration.

Here Known values are

v=0

a=3.20 m/s^{2}

t=32.8s

Unknown value is d

since v=0 , Threrefore we can eliminate vt from the equation

Now substitute these values in above equation,

d=1/2at^{2}

d=1/2(3.20*32.8^{2})

d=1720m rounded to three significant figures

Can you post the equation you are using to plug your variables in? You can edit your question to add more details to it.

Given that

Acceleration a = 3.20\frac{m}{sec^2}.

Time t = 32.8sec

Initial velocity v_i = 0

Suitable equation to calculate the distance.

d=v_it + \frac{1}{2}at^2

where d is distance,v_i is initial velocity, t is time, and a is acceleration.

Sincev_i=0, v_it = 0

Substitute the known values into the equation and solve.

d = \frac{1}{2}at^2

d=\frac{1}{2}3.20\times(32.8)^2

d=\frac{3.20\times1075.84}{2}

Distance d = 1720 m