I need help on these two question 1 A bullet is shot through

I need help on these two question.

1. A bullet is shot through a solid block of wood 30 centimeters (cm) in thickness. It enters the block at a velocity of 300 meters/second (m/s) and leaves at 10 m/s. Assume the bullet travels in a straight line and that it deceleration is proportional to the square of its velocity.

a Derive and solve the initial value problem for velocity of the bullet as a function of time.

b. Find the position of the bullet as a function of time

C. How long does it take the bullet to pass through the block of wood?

d. If a bullet traveling 1 m/s is considered harmless, how thick should the block of wood be to how serve as an effective shield?

e. According to your model, how thick should the block of wood be in order to bring the bullet to rest?

2. A destroyer on the ocean at night suddenly spies te lights of an enemy submarine on the surface 5 miles away. The submarine (a truly amazing machine that is able to go immediately to its cruising depth and immediately reach full speed!) will either stay submerged right where spotted or proceed at full speed in an unknown but constant direction. What path should the destroyer follow to be certain of passing directly over the sub, no matter which option the submarine chooses? From previous studies, the captain of the destroyer knows that the velocity of her ship is 3 times that of the submarine. Be careful with the solution to this problem. You must start the solution the instant that the captain spots the light of the submarine.

Please provide the solutions in step by step.

I would very appreciate your help.

Solution

1) v²-u² = 2as => (10)² - (300)² = 2a(0.3) => 100-30000 = 0.6a => a=-29900/0.6 = -49833.3

Also dv/dt = -kv² =>dv/v² =-kdt => -1/v = -kt+C

v(0) = 300 => -1/300 = C = -0.00333

So we get v(t) = 1/(kt+0.0033)

(b) Now ds/dt = 1/(kt+0.0033) => s= 1/kln(kt+0.0033) +A

s(0) = 0 => 0=1/k ln(0.0033) +A => s(t) = 1/k(ln(kt+0.0033)-ln(0.0033)

(c) s(t) = 0.3 = 1/k ln(kt+0.0033)-ln(0.0033) =>