Answer:
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) [tex]a = \frac{V}{t}[/tex] and (2)[tex]x=x_{0} +V_{0}t + \frac{1}{2} at^2[/tex]
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):
[tex]x=x_{0} +V_{0}t + \frac{1}{2} ( \frac{V}{t})t^2[/tex] We can cancel time because it is mutiplying and dividing the same factor so we have
[tex]x=x_{0} +V_{0}t + \frac{1}{2} Vt[/tex]
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.
[tex]0.210=0 +0*t + \frac{1}{2} 55t[/tex]
[tex]0.210=27.5t[/tex]
[tex]\frac{0.21}{27.5} = t\\[/tex]
t = 0.0076s
a) The acceleration took 0.0076s
Now we replace in the (1) equation the values of time and velocity
[tex]a = \frac{V}{t}[/tex]
[tex]a = \frac{55}{0.0076}[/tex]
a = 7202.4 m/s^2
b) The aceleration was of 7202.4 m/s^2
