A train which is moving with uniform acceleration is observed to take 20 and 30 seconds to travel succesive 0.4km.how much far then will it travel before coming to rest, if the acceleration remains constant
V1 = 400m/20s = 20 m/s.
V2 = 400m/30s = 13.33 m/s.
a = (V2-V1)/(T2-T1) = (13.33-20)/(30-20) = -0.667 m/s^2.
V2 = V1 + 2a*d.
13.33 = 20 - 2*0.667*d. d = ?.
But answer is 162m
V1 = 400m/20s = 20 m/s.
V2 = 400m/30s = 13.33 m/s.
a = (V2-V1)/(T2-T1) = (13.33-20)/(30-20) = -0.667 m/s^2.
V2 = V1 + 2a*d.
13.33 = 20 - 2*0.667*d. d = ?.
To solve this problem, we can use the equations of motion for uniformly accelerated motion. The first equation is:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. Since the train comes to rest, its final velocity (v) will be zero.
So, the equation becomes:
0 = u + at
Rearranging the equation to solve for u, we get:
u = -at
The second equation is:
s = ut + 1/2 at^2
where s is the distance traveled. We know the initial velocity (u), the time taken (t), and the acceleration (a). We need to find the distance traveled (s).
Now, let's calculate the acceleration (a) using the given information. The train takes 20 seconds to travel 0.4 km, so the initial velocity can be calculated as:
u = (0.4 km) / (20 s) = 0.02 km/s
Similarly, the train takes 30 seconds to travel the next 0.4 km, so we can calculate the final velocity as:
v = (0.4 km) / (30 s) = 0.0133333 km/s
Now, we can use the equation v = u + at to calculate the acceleration (a):
a = (v - u) / t = (0 - 0.02 km/s) / 20 s = -0.001 km/s^2
Since we have the acceleration, we can use it to find the distance traveled (s) when the train comes to rest. We know the final velocity (0 km/s) and acceleration (-0.001 km/s^2). Let's substitute these values into the equation s = ut + 1/2 at^2:
0 = (0.02 km/s)(t) + 1/2 (-0.001 km/s^2)(t^2)
Simplifying the equation, we get:
0 = 0.02t - 0.0005t^2
Rearranging the equation, we have:
0.02t = 0.0005t^2
Dividing both sides by t, we get:
0.02 = 0.0005t
Solving for t, we find:
t = (0.02) / (0.0005) = 40 seconds
Now, let's substitute this value of t into the equation s = ut + 1/2 at^2 to find the distance traveled (s):
s = (0.02 km/s)(40 s) + 1/2 (-0.001 km/s^2)(40 s)^2
Simplifying the equation, we get:
s = 0.8 km - 0.8 km = 0 km
Therefore, the train will travel 0 km before coming to rest if the acceleration remains constant.