As per kinematics equation we are given that
[tex]v^2 = v_o^2 + 2ax[/tex]
now we are given that
a = 2.55 m/s^2
[tex]v_0 = 21.8 m/s[/tex]
[tex]v = 0[/tex]
now we need to find x
from above equation we have
[tex]0^2 = 21.8^2 + 2(2.55)x[/tex]
[tex]0 = 475.24 + 5.1 x[/tex]
[tex]x = 93.2 m[/tex]
so it will cover a distance of 93.2 m
Answer:
x = -93.18 meters
Explanation:
The equation is given as :
[tex]v^2=v_o^2+2ax[/tex]...........(1)
The known values are as follows:
[tex]a=2.55\ m/s^2[/tex]
[tex]v_o=21.8\ m/s[/tex]
[tex]v=0\ m/s[/tex]
Putting all the values in equation (1) as :
[tex]0=(21.8\ m/s)^2+2\times 2.55\ m/s^2\times x[/tex]
x = -93.18 meters
So, the vale of x is 93.18 meters. Hence, this is the required solution.