Answer :
(a) The initial velocity is, 14.8 m/s
(b) The acceleration is, [tex]-2.33m/s^2[/tex]
Explanation :
By the 1st equation of motion,
[tex]v=u+at[/tex] ...........(1)
where,
v = final velocity = 0 s
u = initial velocity
t = time = 6.35 s
a = acceleration
The equation 1 will be:
[tex]0=u+a(6.35}[/tex]
[tex]u=-6.35a[/tex] ..........(2)
By the 2nd equation of motion,
[tex]s=ut+\frac{1}{2}at^2[/tex] ...........(3)
where,
s = distance = 47 m
Now substitute equation 2 in 3, we get:
[tex]47=(-6.35a)\times (6.35)+\frac{1}{2}\times a\times (6.35)^2[/tex]
By solving the term, we get:
[tex]a=-2.33m/s^2[/tex]
The acceleration is, [tex]-2.33m/s^2[/tex]
Now we have to calculate the initial velocity.
Using equation 2, we gte:
[tex]u=-6.35a[/tex]
[tex]u=-6.35s\times (-2.33m/s^2)[/tex]
[tex]u=14.8m/s[/tex]
The initial velocity is, 14.8 m/s
