Question

A fireworks 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

Answer 1

a) The acceleration took 0.0076s

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

Step-by-step explanation:

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

(1)
`a = (V)/(t)` and (2)
`x=x_(0) +V_(0)t + (1)/(2) at^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):

`x=x_(0) +V_(0)t + (1)/(2) ( (V)/(t))t^2` We can cancel time because it is mutiplying and dividing the same factor so we have

`x=x_(0) +V_(0)t + (1)/(2) Vt`

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.

`0.210=0 +0*t + (1)/(2) 55t`

`0.210=27.5t`

`(0.21)/(27.5) = t\\`

t = 0.0076s

a) The acceleration took 0.0076s

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

`a = (V)/(t)`

`a = (55)/(0.0076)`

a = 7202.4 m/s^2

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