Answer:
1) d
2) 5 m/s
3) 100
Explanation:
The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:
(i) 
The equation for velocity v and a constant acceleration a is:
(ii) 
1) Solve equation (ii) for acceleration a and plug the result in equation (i)
(iii) 
(iv) 
Simplify equation (iv) and use the given values v = 0, x₀ = 0:
(v) 
2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:
3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

Answer:
gather the data needed to make the forecast
Answer:
a
The radial acceleration is 
b
The horizontal Tension is 
The vertical Tension is 
Explanation:
The diagram illustrating this is shown on the first uploaded
From the question we are told that
The length of the string is 
The mass of the bob is 
The angle made by the string is 
The centripetal force acting on the bob is mathematically represented as

Now From the diagram we see that this force is equivalent to
where T is the tension on the rope and v is the linear velocity
So

Now the downward normal force acting on the bob is mathematically represented as

So

=> 
=> 
The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

=> 
substituting values


The horizontal component is mathematically represented as

substituting value

The vertical component of tension is

substituting value

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

substituting value
![T = [(0.3294) i + (3.3712)j ] \ N](https://tex.z-dn.net/?f=T%20%20%3D%20%5B%280.3294%29%20i%20%20%2B%20%283.3712%29j%20%5D%20%5C%20%20N)