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 ?

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