A train starts from rest at a station and accelerates at the rate of 1.4 m/s2. How long does it take the train to cover a 590 m distance?
v = Vi + a t
x = Xi + Vi t + (1/2) a t^2
Vi = 0, starts at rest
a = 1.4 m/s^2
so
distance = x - Xi = 590 = 0 t + (1/2)(1.4) t^2
0.7 t^2 = 590
t^2 = 843
t = 29 seconds
d = 1/2 a t^2
590 = 1/2 * 1.4 * t^2
t = √(2 * 590 / 1.4)
To find out how long the train takes to cover a distance of 590 m, we can use the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Where:
s = distance covered by the train (590 m)
u = initial velocity of the train (0 m/s, since it starts from rest)
a = acceleration of the train (1.4 m/s^2)
t = time taken by the train
We can rearrange the equation to solve for t:
\[ t = \sqrt{\frac{2s}{a}} \]
Plugging in the given values, we have:
\[ t = \sqrt{\frac{2 \times 590}{1.4}} \]
Now, let's calculate the answer.