A train braking with constant deceleration covers 1km in 20s, and a second kilometre in 30s. Find the Deceleration. Wgat further distance will it cover before coming to a stop. and how long will it take?
This is wrong
To find the deceleration, we'll use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance covered
u = initial velocity
t = time
a = acceleration/deceleration
First, let's find the initial velocity (u). We know that the train is braking, so its initial velocity is zero.
For the first kilometer (s = 1 km, t = 20 s), the equation becomes:
1 = (1/2) * a * (20^2)
Simplifying the equation:
40a = 1
a = 1/40
So, the deceleration is 1/40 km/s^2.
Now, let's find how much further the train will travel before coming to a stop.
To calculate this distance, we need to determine the time it takes for the train to stop. Since we know the initial velocity is zero, we'll use the equation:
v = u + at
Where:
v = final velocity (0 km/s)
u = initial velocity (0 km/s)
a = deceleration
t = time to stop
0 = 0 + (1/40) * t
Simplifying:
t = 0
The time to stop is 0 seconds.
Now, let's find the distance the train will travel before coming to a stop. Since we know the initial velocity is zero and the time is zero, the equation becomes:
s = (1/2) * a * (t^2)
s = (1/2) * (1/40) * (0^2)
s = 0
The train will travel 0 km before coming to a stop.
In summary, the deceleration is 1/40 km/s^2. The train will cover 0 km before coming to a stop, and it will take 0 seconds to stop.
1km = 1000m.
Vo = 1000m/20s = 50m/s.
Vf = 1000m/30s = 33.33m/s.
a = (Vf-Vo)/t = (33.33 - 50)/(30-20) =
-16.67 / 10 = -1.67m/s^2.
The negative sign meane deceleration.
Vf^2 = Vo^2 + 2ad = 0,
(33.33)^2 + 2*(-1.67)d = 0,
1111.1 -3.34d = 0,
-3.34d = -1111.1,
d = 332.7m.
Vf = Vo + at,
t = (Vf-Vo)/a,
t = (0 - 33.33) / -1.67 = 20s.