A train accelerates uniformly from 30 km/hour to 60 km/hour in 30 seconds.
What are the acceleration anddistance covered?
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
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.
oops. Go with Henry2
I forgot to include the initial velocity.
To find the acceleration of the train, we can use the formula:
acceleration = (change in velocity) / (time)
Given that the initial velocity (u) is 30 km/hour and the final velocity (v) is 60 km/hour, the change in velocity is:
change in velocity = v - u
= 60 km/hour - 30 km/hour
= 30 km/hour
The time taken (t) is given as 30 seconds.
Now, we can plug in the values into the formula to find the acceleration:
acceleration = (change in velocity) / (time)
= 30 km/hour / 30 seconds
= 1 km/hour^2
Therefore, the acceleration of the train is 1 km/hour^2.
To find the distance covered (s), we can use the formula:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Plugging in the values:
distance = (30 km/hour * 30 seconds) + (0.5 * 1 km/hour^2 * (30 seconds)^2)
= (900 km) + (0.5 * 1 km/hour^2 * 900 seconds^2)
= 900 km + 0.5 km/hour^2 * 81000 seconds
= 900 km + 0.5 km/hour^2 * 81000 seconds
= 900 km + 40500 km
= 41400 km
Therefore, the distance covered by the train is 41400 km.