Sindhuja Parimi
3

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


d=vt+1/2at2


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/s2

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=1/2at2

d=1/2(3.20*32.82)

d=1720m rounded to three significant figures

Mahesh Godavarti
1

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

Krishna
1

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