A train driver moving at 40km|h applies break so that there is a constant retardation of 0.5m|s find time taken before the train stops

Vo = 40.000m/3600s = 11.11 m/s.

V = Vo + a*t = 0, 11.11 - 0.5t = 0, -0.5t = -11.11, t = 22.22 s.

To find the time taken before the train stops, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s, as the train stops)
u = initial velocity (40 km/hr = 40,000 m/3600 s = 11.11 m/s, as the train is moving)
a = acceleration (retardation) = -0.5 m/s^2 (negative sign as it is opposing motion)
s = distance traveled

Rearranging the equation, we get:

0 = (11.11)^2 + 2(-0.5)s

Simplifying:

0 = 123.21 - s

s = 123.21 m

Now we can use the equation of motion to find the time taken:

v = u + at

0 = 11.11 + (-0.5)t

-11.11 = -0.5t

t = -11.11 / -0.5

t = 22.22 seconds

Therefore, the time taken before the train stops is 22.22 seconds.

To find the time taken before the train stops, we can use the kinematic equation:

\(v^2 = u^2 + 2as\)

Where:

\(v\) = final velocity (0 m/s, as the train stops)
\(u\) = initial velocity (40 km/h converted to m/s)
\(a\) = acceleration (retardation in this case)
\(s\) = distance traveled

Let's break down the given information:

Initial velocity, \(u\) = 40 km/h
Final velocity, \(v\) = 0 m/s
Acceleration, \(a\) = -0.5 m/s² (negative sign due to retardation)

First, we need to convert the initial velocity from km/h to m/s:

1 km = 1000 m (conversion factor)
1 hour = 3600 seconds (conversion factor)

\(u = 40 \times \frac{1000}{3600} = \frac{40000}{3600} = \frac{100}{9}\) m/s

Now, let's substitute the given values into the equation:

\(0^2 = \left(\frac{100}{9}\right)^2 + 2 \times (-0.5) \times s\)

Simplifying the equation:

\(0 = \frac{10000}{81} - s\)

Rearranging the equation to isolate \(s\):

\(s = \frac{10000}{81}\)

Finally, we know that distance, \(s\), is given by:

\(s = ut + \frac{1}{2}at^2\)

Since the train stops, the distance \(s\) is equal to 0. Therefore, we can solve for \(t\) using the given values:

\(0 = \frac{100}{9}t + \frac{1}{2} \times (-0.5) \times t^2\)

Rearranging the equation:

\(0 = \frac{100}{9}t - 0.25t^2\)

Multiplying both sides by 9:

\(0 = 100t - 2.25t^2\)

This equation is a quadratic equation. To solve it, we can re-write it in the standard form:

\(2.25t^2 - 100t = 0\)

Factoring out \(t\):

\(t(2.25t - 100) = 0\)

Setting each factor to zero and solving for \(t\):

\(t = 0\) or \(2.25t - 100 = 0\)

From the second equation, we get:

\(2.25t = 100\)

Dividing both sides by 2.25:

\(t = \frac{100}{2.25} = \frac{400}{9}\)

Therefore, the time taken before the train stops is \(t = \frac{400}{9}\) seconds.