Problema Solution
A train is traveling at a rate of 32 km/hr and later at the bottom of a grade 40 seconds later is going 80 km/hr. What is its acceleration in km/hr^2? What is the average velocity down the grade? How long is the grade?
Answer provided by our tutors
v0 = 32 km/h the initial velocity
v1 = 80 km/h the final velocity
t = 40 s = 40/3600 hr = 1/90 hr
a = (v1 - v0)/t is the formula for the acceleration
a = (80 - 32)/(1/90)
a = 4,320 km/hr^2
the length of the grade d = vot + (1/2)a*t^2
d = 32*(1/90) + (1/2)*4320*(1/90)^2
d = 28/45
d = 0.62 km
click here to see the step by step calculation for d:
the average velocity v = d/t
v = (28/45)/(1/90)
v = 56 km/h
another way to find the velocity is to find the average of the initial and the final velocity:
v = (1/2)(v0 + v1)
v = (1/2)(32 + 80)
v = 56 km/h.