Question

A train moving with a velocity of 88.1 km/hour North, increases its speed with a uniform acceleration of 0.250 m/s2 North until it reaches a velocity of 160.0 km/hour North. What distance did the train travel while it was increasing its velocity, in units of meters? Give the answer as a positive number

Answer 1

Given :-

Initial Velocity of Train = u = 88.1 Km/h = 24.4 ms-¹

Final velocity of train = v = 160 Km/h = 44.4 ms-¹

Acceleration of Train = a = 0.25 ms-²

Using First Equation of Motion,

v - u = at

44.4 - 24.4 = 0.25t

t = 20/0.25

t = 80 s

Again,

Distance traveled,

S = ut + 1/2 at²

S = (24.4)(80) + 1/2(0.25)(80)(80)

S = 1952 + 800

S = 2752 m

Finding distance traveled using third equation of Motion,

v² - u² = 2aS

(44.4)² - (24.4)² = 2(0.25)S

1971.36 - 595.36 = 0.5S

S = 1376/0.5

S = 2752 m

Hence,

Distance travelled = S = 2752 m