a = ∆v/∆t = (30km/hr)/30s = 1km/hr/s = 3600 km/hr^2
As usual, s = 1/2 at^2 = 1/2 * 3600 * (30/3600)^2 = 1/8 km
Or, you might want to use
30km/hr/s * 1000m/km * 1hr/3600s = 0.2778 m/s^2
s = 1/2 * 0.2778 * 30^2 = 125 m
A train accelerates uniformly from 30 km/hour to 60 km/hour in 30 seconds.
What are the acceleration anddistance covered?
4 answers
thanks.
Vo = 30km/h = 30,000m/3600s. = 8.33 m/s.
V = 60km/h = 60,000m/3600s = 16.67 m/s.
V = Vo + a*t = 16.67.
8.33 + a*30 = 16.67,
a = 0.278 m/s^2.
V^2 = Vo^2 + 2a*d = 16.67^2.
8.33^2 + 0.555d = 16.67^2,
0.555d = 208.5,
d = 376 m.
V = 60km/h = 60,000m/3600s = 16.67 m/s.
V = Vo + a*t = 16.67.
8.33 + a*30 = 16.67,
a = 0.278 m/s^2.
V^2 = Vo^2 + 2a*d = 16.67^2.
8.33^2 + 0.555d = 16.67^2,
0.555d = 208.5,
d = 376 m.
oops. Go with Henry2
I forgot to include the initial velocity.
I forgot to include the initial velocity.
1 answer