a train's stopping distance even when full emergency brakes are engaged is 1.3 km. If the train was travelling at an initial velocity of 90km/h(forward), determine its acceleration under full emergency braking
To determine the acceleration of the train under full emergency braking, we need to use the equation of motion:
v² = u² + 2as
Where:
v = final velocity (0 m/s, since the train comes to a stop)
u = initial velocity (90 km/h = 25 m/s)
a = acceleration
s = stopping distance (1.3 km = 1300 m)
Now, we can rearrange the equation to solve for acceleration (a):
a = (v² - u²) / (2s)
Substituting the given values, we have:
a = (0 - 25²) / (2 * 1300)
Calculating this expression, we get:
a ≈ -15625 / 2600
Simplifying the fraction, we have:
a ≈ -6.02 m/s²
Therefore, the acceleration of the train under full emergency braking is approximately -6.02 m/s² (negative sign indicates deceleration).
km/h -> m/s
km -> m
a=v²/2s