V = u + at where u is initial velocity (15 m/s), a is acceleration (2m/s^2) and t is time (15 seconds)
V = 15 + 2 X 15
V = 45 m/s
Complete Question
The complete question is shown on the first uploaded image
Answer:
a

b
New 
Explanation:
From the question we are told that
The refractive index of the core is 
The refractive index of the cladding is 
Generally according to Snell's law

Where
is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection
![\theta_{max} = 90 - sin^{-1} [\frac{n_{cladding}}{n_{core}} ]](https://tex.z-dn.net/?f=%5Ctheta_%7Bmax%7D%20%3D%2090%20-%20sin%5E%7B-1%7D%20%5B%5Cfrac%7Bn_%7Bcladding%7D%7D%7Bn_%7Bcore%7D%7D%20%5D)
![\theta_{max} = 90 - sin^{-1} [\frac{1.421}{1.497}} ]](https://tex.z-dn.net/?f=%5Ctheta_%7Bmax%7D%20%3D%2090%20-%20sin%5E%7B-1%7D%20%5B%5Cfrac%7B1.421%7D%7B1.497%7D%7D%20%5D)

Given from the question the the largest angle is 5°
Generally the refraction index of the cladding is mathematically represented as


The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.
2. This is immediate from (2.2.7).
3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =
To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as
F = ma
Where,
m= Mass
a = Acceleration
At the same time the frictional force can be defined as,

Where,
Frictional coefficient
N = Normal force (mass*gravity)
Our values are given as,

By condition of Balance the friction force must be equal to the total net force, that is to say



Re-arrange to find acceleration,



Therefore the acceleration the horse can give is 