A train travels 30m/s then breaks are suddenly applied producing a retarding force equal to 1/10 of the weight of the train runs on the distance through which the train runs on the track before coming to rest in.
F = - mg/10 = m a
so
a = -g/10
v = Vi + at = 30 - g t /10
at stop v = 0
t = 300/g = time to stop
d = Vi t + (1/2)a t^2
= 30(300/g) -(1/2)(g/10)(300^2/g^2)
= 9000/g - (1/2)(9000/g)
= 4500/g
about 450 meters
alternate way using energy
(1/2) m Vi^2 = F d
F = mg/10
(1/2) m Vi^2 = m g d/10
(30)^2 = g d/5
900 *5 /g = d
d = 4500/g
To find the distance through which the train runs before coming to rest, we'll need to use the equations of motion. The braking force is given as 1/10 of the weight of the train.
Let's assume the weight of the train is "W" newtons.
1. Find the braking force (F):
F = (1/10)W
2. Calculate the acceleration (a):
Since the force of friction acts in the opposite direction of motion, it causes deceleration.
F = ma (Newton's second law)
(1/10)W = ma (Substitute the value of F)
a = (1/10)W / m (Divide both sides by mass)
3. Calculate the time taken to stop (t):
Initial velocity (u) = 30 m/s (given)
Final velocity (v) = 0 m/s (since it comes to rest)
a = (v - u) / t (one of the equations of motion)
(1/10)W / m = (0 - 30) / t (Substitute the values of a, v, and u)
t = -10m / W (Rearrange the equation)
4. Calculate the distance traveled (s):
We can use another equation of motion: s = ut + (1/2)at^2
s = 30t + (1/2)(1/10)W / m * t^2 (Substitute the values)
s = 30t + (1/20)(W / m)t^2
Therefore, the distance through which the train runs before coming to rest is given by the equation:
s = 30t + (1/20)(W / m)t^2
To find the distance through which the train runs on the track before coming to rest, we can use the concept of deceleration and the equation of motion.
First, let's calculate the deceleration (retarding force) experienced by the train. The retarding force is equal to 1/10 of the weight of the train. Let's denote the weight of the train as W.
Retarding force = 1/10 * W
Next, we need to find the mass of the train. The weight of an object is equal to its mass multiplied by the acceleration due to gravity. Let's denote the mass of the train as M.
W = M * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2)
Therefore, M = W / g
Now, let's substitute the value of the retarding force in terms of the mass of the train:
Retarding force = 1/10 * (M * g)
Next, we can use Newton's second law, F = ma, where F is the retarding force and a is the deceleration:
Retarding force = M * a
Setting both equations equal:
M * a = 1/10 * (M * g)
Now, let's solve for the deceleration, a:
a = 1/10 * g
Now that we know the deceleration, we can use the equation of motion to find the distance through which the train runs on the track before coming to rest.
The equation of motion for uniformly decelerated motion is:
v^2 = u^2 - 2as
Where:
v = final velocity (in this case, 0 m/s as the train comes to rest)
u = initial velocity (given as 30 m/s)
a = deceleration
s = distance
Substituting the known values:
0^2 = (30)^2 - 2 * a * s
Simplifying:
900 = 2 * a * s
Dividing both sides by 2a:
s = 900 / (2 * a)
Now, substitute the value of deceleration, a:
s = 900 / (2 * (1/10 * g))
Simplifying further:
s = 900 / (2 * (1/10 * 9.8))
s = 900 / (2 * 0.98)
s = 900 / 1.96
s ≈ 459.18 meters
Therefore, the train runs on a distance of approximately 459.18 meters on the track before coming to rest.