The velocity of a train is reduced uniformly from 40km/sec to 30km/sec travelling a distance of 100km.

A. Calculate the retardation
B. How much farther will it travel before coming to rest, assuming the same retardation.

how long did it take?

40t + 1/2 at^2 = 100
40 + at = 30
Solve that to find a and t.

Now find another t such that 30 + at = 0
Note that a is in km/s^2

A. V^2 = Vo^2 + 2a*d.

30^2 = 40^2 + 2a*100.
900 - 1600 = 200a,
a = -3.5km/s^2.

B. V^2 = Vo^2 + 2a*d.
0 = 40^2 + (-7)*d,
d = __km.

d - 100km = __km farther.

A. To calculate the retardation, we need to find the change in velocity and the distance traveled.

Change in velocity (Δv) = Final velocity (v_f) - Initial velocity (v_i)
= 30 km/sec - 40 km/sec
= -10 km/sec (Note: the negative sign indicates reduction)

Distance traveled (s) = 100 km

Retardation (a) = Δv / s
= (-10 km/sec) / (100 km)
= -0.1 km/sec²

Therefore, the retardation is -0.1 km/sec² (negative sign indicates reduction).

B. To find how much farther the train will travel before coming to rest, assuming the same retardation, we need to calculate the distance it will cover when its velocity becomes zero.

We can use the equation:

Final velocity (v_f) = Initial velocity (v_i) + (acceleration * time)

Since we know the final velocity will be zero (v_f = 0), we can solve for time:

0 = 40 km/sec + (-0.1 km/sec²) * time

Rearranging the equation:

-40 km/sec = -0.1 km/sec² * time

Dividing both sides by -0.1 km/sec²:

400 = time

Therefore, it will take 400 seconds for the train to come to rest.

To find the distance traveled (d) during this time, we can use the equation:

distance (d) = (initial velocity * time) + (0.5 * acceleration * time²)

Plugging in the values:

d = (40 km/sec * 400 sec) + (0.5 * -0.1 km/sec² * (400 sec)²)

Calculating:

d = 16,000 km + 8,000 km

d = 24,000 km

Therefore, the train will travel 24,000 km farther before coming to rest, assuming the same retardation.

To calculate the retardation, we can use the formula:

Retardation (a) = (Final Velocity (v) - Initial Velocity (u)) / Time (t)

In this case, the initial velocity is 40 km/sec, the final velocity is 30 km/sec, and the distance traveled is 100 km. We need to find the time taken (t) first.

To find the time taken, we can use the formula:

Distance (d) = (Initial Velocity (u) + Final Velocity (v)) / 2 * Time (t)

Substituting the known values, we have:

100 km = ((40 km/sec + 30 km/sec) / 2) * t

Simplifying further:

100 km = (70 km/sec / 2) * t
200 km = 70 km/sec * t
t = 200 km / 70 km/sec
t ≈ 2.857 seconds

Now, we can substitute the values into the retardation formula:

Retardation (a) = (30 km/sec - 40 km/sec) / 2.857 sec
Retardation (a) = -10 km/sec / 2.857 sec
Retardation (a) ≈ -3.5 km/sec²

Therefore, the retardation of the train is approximately -3.5 km/sec².

To calculate how much farther the train will travel before coming to rest, we need to determine the time it would take for the train to come to rest. Since the retardation remains the same, we can use the formula:

Retardation (a) = (Final Velocity (v) - Initial Velocity (u)) / Time (t)

Rearranging the formula to solve for time:

Time (t) = (Final Velocity (v) - Initial Velocity (u)) / Retardation (a)

Given that the final velocity (v) is 0 km/sec, the initial velocity (u) is 30 km/sec, and the retardation (a) is -3.5 km/sec², we can calculate the time taken for the train to come to rest:

Time (t) = (0 km/sec - 30 km/sec) / -3.5 km/sec²
Time (t) = -30 km/sec / -3.5 km/sec²
Time (t) ≈ 8.571 seconds

To find the distance traveled in this time, we can use the formula:

Distance (d) = Initial Velocity (u) * Time (t) + (1/2) * Retardation (a) * Time (t)^2

Substituting the known values:

Distance (d) = 30 km/sec * 8.571 sec + (1/2) * (-3.5 km/sec²) * (8.571 sec)^2

Simplifying further:

Distance (d) ≈ 257.13 km + (1/2) * (-3.5 km/sec²) * 73.34 sec²
Distance (d) ≈ 257.13 km - 89.52 km
Distance (d) ≈ 167.61 km

Therefore, the train will travel approximately 167.61 km farther before coming to rest, assuming the same retardation.