Answer :
(a) The initial velocity is, 14.8 m/s
(b) The acceleration is,
![-2.33m/s^2](https://img.qammunity.org/2020/formulas/physics/college/ganvwq3b6t263e8ucs4cuowrczgn1vs9c4.png)
Explanation :
By the 1st equation of motion,
...........(1)
where,
v = final velocity = 0 s
u = initial velocity
t = time = 6.35 s
a = acceleration
The equation 1 will be:
![0=u+a(6.35}](https://img.qammunity.org/2020/formulas/physics/college/cao5fzx8ryjffpi5s1i6hojoqxij2ubepw.png)
..........(2)
By the 2nd equation of motion,
...........(3)
where,
s = distance = 47 m
Now substitute equation 2 in 3, we get:
![47=(-6.35a)* (6.35)+(1)/(2)* a* (6.35)^2](https://img.qammunity.org/2020/formulas/physics/college/zg07czix1qwfm38p3jqtrcp6hako28xfqw.png)
By solving the term, we get:
![a=-2.33m/s^2](https://img.qammunity.org/2020/formulas/physics/college/pa93vpqzw5vo16lw813a881gtzvl530j96.png)
The acceleration is,
![-2.33m/s^2](https://img.qammunity.org/2020/formulas/physics/college/ganvwq3b6t263e8ucs4cuowrczgn1vs9c4.png)
Now we have to calculate the initial velocity.
Using equation 2, we gte:
![u=-6.35a](https://img.qammunity.org/2020/formulas/physics/college/nkl4ewiwcwda45s2na90io44f3hm0t4gza.png)
![u=-6.35s* (-2.33m/s^2)](https://img.qammunity.org/2020/formulas/physics/college/ecgohvzkfytjdqsz42x64btr6xc2ohxezu.png)
![u=14.8m/s](https://img.qammunity.org/2020/formulas/physics/college/k9owh8hrtlvf2ja8ydtw5jtwlcowwum711.png)
The initial velocity is, 14.8 m/s