Question

At t=​0, a train approaching a station begins decelerating from a speed of 140 mi/hr according to the acceleration function a(t)= -1120 (1+4t)^-3 mi/hrsquared2​,where t≥0. How far does the train travel between t=0 and t=0.1? Between t=0.1 and t=0.3​?

Answer 1

To find the distance traveled by the train between t=0 and t=0.1 and t=0.1 and t=0.3, we need to integrate the velocity function. The velocity function is obtained by integrating the given acceleration function. After finding the velocity function, we can integrate it to find the distances traveled.

Step-by-step explanation:

The question asks for the distance traveled by the train between two given times, t=0 and t=0.1 and t=0.1 and t=0.3. To find the distance, we need to integrate the velocity function. The velocity function is the integral of the acceleration function, so we'll start by finding the velocity function.

Given:

`a(t) = -1120 (1+4t)^(-3) mi/hrs^2`

To find the velocity function, we'll integrate a(t) with respect to t.

v(t) = ∫a(t)dt

To integrate the function, we'll use the power rule and the constant multiple rule. After integrating, we'll evaluate the integral at the upper and lower limits of t to find the velocities.

Using calculus, we find that the velocity function is:

`v(t) = -280(1+4t)^(-2) + C`

Next, to find the distance traveled, we integrate the velocity function over the given time intervals.

Distance between t=0 and t=0.1:

d1 = ∫[0,0.1] v(t) dt

Similarly, the distance between t=0.1 and t=0.3 can be calculated.

Answer 2

9.04 miles

7.24 miles

Step-by-step explanation:

Using

S =[ (V+U)/2]t₀........................ Equation 1

Where S = distance, V = final velocity, U = initial velocity, t = time.

Between t = 0 and t = 0.1

U = 140 mi/hr, t₀ = 0.1 hr and

V = at, where a = acceleration

a = 1120(1+4t)⁻³ at t = 0.1 ( as the train is decelerating)

a = 1120(1+4×0.1)⁻³

a = 1120(1.4)⁻³ = -1120/1.4³

a = 408.16 mi/hr²

Therefore, V = 408.16(0.1)

V = 40.82 mi/hr.

Substituting these values into equation 1,

S = {(40.82+140)/2}0.1

S = 9.04 miles

Between t = 0.1 and t = 0.3,

U = 40.82 mi/hr, t₀ = 0.3 - 0.1 = 0.2 hr, v = at

a = -1120(1+4t)⁻³ at t = 0.3

a = 1120/[1+4(0.3)]³

a = 1120(2.2)³

a =105.18 mi/hr

V = 105.18×0.3

V = 31.55 mi/hr.

Also substituting into equation 1

S = [(31.55+40.82)/2]0.2

S = 7.24 miles