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

Answer:

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