A train travels at a speed of 24 m/s. Then it slows down uniformly at 0.065 m/s2 until it stops. What distance does the train travel while slowing down?
v^2 = u^2 - 2*a*s
here, v = 0 (final speed)
u = 24 m/s
a = 0.065 m/s^2
Solve above to obtain s - distance covered while slowing down.
To determine the distance the train travels while slowing down, you need to use the equations of motion.
First, you need to find the time it takes for the train to come to a stop. To do this, you can use the equation:
v = u + at
where:
v is the final velocity (0 m/s),
u is the initial velocity (24 m/s),
a is the acceleration (-0.065 m/s^2),
and t is the time.
Rearranging the equation, we get:
t = (v - u) / a
Plugging in the values, we have:
t = (0 - 24) / -0.065
Simplifying, we get:
t = 369.23 seconds (rounded to two decimal places)
Now that we have the time it takes for the train to stop, we can find the distance traveled during this time using the equation:
s = ut + (1/2)at^2
where:
s is the distance,
u is the initial velocity (24 m/s),
t is the time (369.23 s),
a is the acceleration (-0.065 m/s^2).
Plugging in the values, we have:
s = (24 × 369.23) + (1/2) × (-0.065) × (369.23)^2
Simplifying, we get:
s ≈ 5323.09 meters
Therefore, the train travels approximately 5323.09 meters while slowing down.