Question

a plane averaged 490 mph on a trip going east from springfield to shelbyville, but only 310 mph on the return trip from shelbyville back to springfield. the total flying time in both directions was (t = 11.5) hrs. let (k) be the proportion of the total flight time (t), so that the flight from springfield to shelbyville takes (kt) hours, where (k) is between 0 and 1. 1a write an expression for the duration of the return trip from shelbyville to springfield in terms of (k) and (t). 1b write an equation for the one - way distance (d) from springfield to shelbyville in terms of (k) and (t).

Explanation:

Step1: Usar fórmula de distancia

La distancia $d$ se calcula como $d = v\times t$, donde $v$ es la velocidad y $t$ es el tiempo. La velocidad del viaje de Springfield a Shelbyville es $v = 490$ mph y el tiempo es $kt$.

Step2: Escribir la ecuación

Sustituyendo en la fórmula $d = v\times t$, obtenemos $d=490kt$.

Answer:

$d = 490kt$