A train is moving with 72 kmh−1speed. The driver applies the brakes and the train decelerates with 1.3 ms−2and takes 25 seconds before it stops. Calculate the distance covered by the train after applying the brakes

Question

A train is moving with 72 kmh−1speed. The driver applies the brakes and the train decelerates with 1.3 ms−2and takes 25 seconds before it stops. Calculate the distance covered by the train after applying the brakes

Answer 1

Given values are

vi=72kmsec−1=20msec−1
vf=0ms-1
a=−1.3msec−2
t=25sec
Use second equation of motion to find traveled distance before stopping.
S=Vit+1/2 at2
S=(20)x(25)+1/2(-0.3)(25)2
S=93.75m

Answer 2

Vi = 72000 meters /3600 seconds = 20 m/s

Vf = 0
a = -1.3 m/s^2
v = Vi + a t
0 = 20 -1.3 t
t = 15.4 seconds , Not 25 seconds
d = average speed * time = 10 m/s * 15.4 s = 154 meters

Answer 3

To calculate the distance covered by the train after applying the brakes, we can use the equation for distance covered during deceleration:

distance = initial velocity * time + 0.5 * acceleration * time^2

Given:
Initial velocity (u) = 72 km/h = 72 * (1000/3600) m/s ≈ 20 m/s
Deceleration (a) = -1.3 m/s^2 (negative sign indicates deceleration)
Time (t) = 25 seconds

Using the equation for distance covered during deceleration:

distance = 20 m/s * 25 s + 0.5 * (-1.3 m/s^2) * (25 s)^2

Calculating the equation:

distance = 500 m - 16.25 m/s^2 * 625 s^2
distance ≈ 500 m - 10156.25 m
distance ≈ -9645.25 m

Since the distance covered cannot be negative, we can consider the magnitude of the distance:

distance ≈ |-9645.25 m| ≈ 9645.25 m

Therefore, the train covers approximately 9645.25 meters (or 9.64525 kilometers) before coming to a stop after applying the brakes.

Answer 4

To calculate the distance covered by the train after applying the brakes, we can use the equation:

distance = initial velocity * time + (1/2) * acceleration * time^2

Here, the initial velocity (vi) is given as 72 km/h, which we need to convert to m/s.

To convert from km/h to m/s, we need to divide the speed by 3.6:

Initial velocity (vi) = 72 km/h ÷ 3.6 = 20 m/s

The acceleration (a) is given as -1.3 m/s^2, as it is decelerating. The negative sign indicates the direction of deceleration.

The time (t) taken for the train to stop is given as 25 seconds.

Now, we can substitute the values into the equation:

distance = 20 m/s * 25 s + (1/2) * (-1.3 m/s^2) * (25 s)^2

Simplifying the equation:

distance = 500 m + (1/2) * (-1.3 m/s^2) * 625 s^2

distance = 500 m - 406.25 m

distance = 93.75 m

Therefore, the train covers a distance of 93.75 meters after applying the brakes.