Afireworks shell is accelerated from rest to a velocity of 55.0 m/s over a distance of 0.210 m. (a) how long (in s) did the acceleration last? s
(b) calculate the acceleration (in m/s2). (enter the magnitude.) m/s^2

a) The acceleration took 0.0076s

b) The aceleration was of 7202.4 m/s^2

Explanation:

We need to use the formulas for acceleration movement in straight line that are:

(1)    and  (2)

Where

a = acceleration

V = Velocity reached

Vo = Initial velocity

t = time

x = distance

xo = initial distance.

We have the following information:

a = We want to find      V = 55.0 m/s      

Vo = 0m/s because it starts from rest       t = we want to find      

x = 0.210 m         xo= 0 m we beging in the point zero.

We have to variables in two equations, so we are going to replace in the second equation (2) the aceleration of the first one(1):

We can cancel time because it is mutiplying and dividing the same factor so we have

In this equation we just have one variable that we don't know that is time, so first we are going to replace the values and after that clear time.

t = 0.0076s

a) The acceleration took 0.0076s

Now we replace in the (1) equation the values of time and velocity

a = 7202.4 m/s^2

b) The aceleration was of 7202.4 m/s^2

. west