A motor boat is traveling at a speed of 2.4 m/s shuts off its engine at t=0 . How far does it travel before coming to rest after 3 seconds it's speed have dropped to half of its original value ? Assume that the drag force of the water is proportional to v ( R = -bv)

To find the distance traveled by the motorboat before coming to rest, we need to calculate the time it takes for the boat's speed to decrease to half of its original value, and then use this time to find the distance traveled.

Step 1: Calculate the time it takes for the boat's speed to decrease to half of its original value.
We know that the drag force acting on the boat is proportional to its velocity, given by the equation R = -bv. Since the drag force acts in the opposite direction of the boat's motion, it leads to a decrease in the boat's speed.

Let's assume the initial velocity of the boat is v0 and the deceleration is a (negative value since it's decreasing). The drag force is given by R = -bv, which can be written as ma = -bv, where m is the mass of the boat.

Now, rearranging the equation, we get ma = -bv.
If we divide both sides by m, we have:
a = -(b/m)v

Since acceleration is defined as the rate of change of velocity over time, we can express it as dv/dt. So, now we have:
dv/dt = -(b/m)v

We can separate variables to solve this differential equation:
dv/v = -(b/m)dt
Integrating both sides, we get:
ln(v) = -(b/m)t + C
Where C is the constant of integration.

Since we are given that the boat's speed has dropped to half of its original value, we can write:
v/2 = v0 * e^(-(b/m)t)
Where v0 is the initial velocity of the boat, e is Euler's number (approximately 2.71828), and t is the time it takes for the boat's speed to decrease.

Now, rearranging the equation, we get:
t = -[(m/b) * ln(2)] / (v0)

Step 2: Calculate the distance traveled by the boat.
The formula to calculate distance is given by d = v0t + (1/2)at^2, where v0 is the initial velocity, t is the time, and a is the acceleration.

Since the boat is coming to rest, its final velocity is 0, and the acceleration is the negative value of the initial velocity divided by the time:
a = -(v0/t)

Substituting the value of a and rearranging the equation, we have:
d = v0t - (1/2)(v0/t)t^2
Simplifying further, we get:
d = v0t - (1/2)v0t

The distance traveled by the boat is given by:
d = (1/2)v0t

Now, substituting the value of t from Step 1, we can calculate the distance traveled by the boat:
d = (1/2)v0 * [ -[(m/b) * ln(2)] / (v0) ]

Simplifying, we get:
d = -[(m/b) * ln(2)] / 2
which is the distance traveled by the boat before coming to rest after the time t.