A car accelerates from rest at the rate of 1.4 + 3 (root)t mph per second for 4 seconds. How far does the car travel in those 4 seconds?

To find the distance the car travels in 4 seconds, we first need to calculate the car's velocity as a function of time. We are given that the car's acceleration is given by the equation a(t) = 1.4 + 3√t, and we want to find the distance it travels in 4 seconds.

To calculate the velocity, we need to integrate the acceleration function with respect to time. Let's break down the integration step by step:

∫ (1.4 + 3√t) dt = ∫ 1.4 dt + ∫ 3√t dt

Integrating 1.4 with respect to t gives us: 1.4t

For the second term, let's use u-substitution, where u = √t.
Taking the derivative of both sides: du/dt = 1/2√t
Rearranging, we have: dt = 2√t du

Replacing dt in the original equation: ∫ 3√t dt = ∫ 3(u)(2√t) du = 6u√t du

Now our equation becomes: 1.4t + 6u√t du

Let's integrate the remaining terms:

∫ 1.4t dt = 1.4(1/2)t^2 = 0.7t^2

∫ 6u√t du = 6u(2/3)u^(3/2) = 4u^(5/2)

Now we have the velocity as a function of time: v(t) = 0.7t^2 + 4u^(5/2)

To find the distance traveled in the given time interval (0 to 4 seconds), we need to integrate the velocity function once again:

∫ v(t) dt = ∫ (0.7t^2 + 4u^(5/2)) dt

Integrating 0.7t^2 gives us: (0.7/3)t^3

For the second term, let's integrate with respect to u: ∫ 4u^(5/2) du = (4/(7/2 + 1))u^(7/2 + 1) = (8/9)u^(9/2)

Now replacing u with √t: (8/9)(√t)^(9/2) = (8/9)(√t^9) = (8/9)t^(9/4)

The distance traveled in the given time interval is the definite integral of the velocity function from 0 to 4:

∫[0 to 4] (0.7t^2 + 4u^(5/2)) dt = [(0.7/3)t^3 + (8/9)t^(9/4)] from 0 to 4

Evaluating this expression will give us the distance traveled by the car in 4 seconds.