The speed of the train is reduced from 36 km per hour to 9 km per hour whilst it travels a distance of 150 metres,if the retardation be uniform,find how much further it will travel before coming to rest
Or:
d = -(10^2)/-0.6250 = 160 m.
160 - 150 = 10 m. Further.
Vo = 36km/h = 36,000m/3600s = 10 m/s.
V = 9km/h = 9000m/3600s = 2.5 m/s.
a = (V^2-Vo^2)/2d
a = (2.5^2-10^2)/300 = -0.3125 m/s^2.
d = (V^2-Vo^2)/2a
d = (0-2.5^2)/-0.6250 = 10 m.
d = -(10^2)/-0.6250 = 160 m.
160 - 150 = 10 m. Further.
To find how much further the train will travel before coming to rest, we can use the equation of motion that relates the initial velocity, final velocity, acceleration (which is negative in this case), and distance.
The equation we can use is:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 in this case because the train comes to rest)
u = initial velocity (36 km/h converted to m/s)
a = acceleration (which is the retardation in this case)
s = distance
First, let's convert the initial velocity from km/h to m/s:
1 km/h = (1000 m / 1 km) * (1 h / 3600 s) = 5/18 m/s
So, the initial velocity (u) is:
u = 36 km/h * (5/18) m/s = 10 m/s
Next, we need to determine the acceleration (a). Since the retardation is uniform, it has the same magnitude as the acceleration but in the opposite direction. So, the acceleration is -a.
Now, using the equation of motion and substituting the known values, we have:
0 = (10 m/s)^2 + 2 * -a * 150 m
Simplifying the equation further:
0 = 100 m^2/s^2 - 300a
Rearranging the equation to solve for a:
300a = 100 m^2/s^2
a = (100 m^2/s^2) / 300
a = 1/3 m^2/s^2
Now that we know the acceleration, we can use the equation of motion to find the distance (s) the train will travel before coming to rest.
0 = (10 m/s)^2 + 2 * (1/3 m^2/s^2) * s
0 = 100 m^2/s^2 + (2/3) * s
Simplifying:
(2/3) * s = -100 m^2/s^2
s = (-100 m^2/s^2) / (2/3)
s = -150 m^2/s^2
Since distance cannot be negative, we take the absolute value of s:
s = |-150 m^2/s^2| = 150 m^2/s^2
Therefore, the train will travel an additional 150 meters before coming to rest.
Vo = 36km/h = 36,000m/3600s = 10 m/s.
V = 9km/h = 9000m/3600s = 2.5 m/s.
a = (V^2-Vo^2)/2d
a = (2.5^2-10^2)/300 = -0.3125 m/s^2.
d = (V^2-Vo^2)/2a
d = (0-2.5^2)/-0.6250 = 10 m. Further.