Answer:
d = 493.72 m
Explanation:
Given that,
Initial velocity of the train, u = 80 km/h = 22.22 m/s
Acceleration of the train, a = -0.5 m/s² (negative as it slows down)
Finally brakes are applied, v = 0
We need to find the distance the train travels. Let the distance be d. Using third equation of kinematics to find it.
So, the required distance is equal to 493.72 m.
Answer:
Speed: 4.8 m/s
Velocity: 4.8 m/s north
Explanation:
Definitions:
- Speed is a scalar quantity, which is equal to the ratio between distance covered (d) and time taken (t):
- Velocity is a vector quantity, whose magnitude is equal to the ratio between the displacement of the object and the time taken:
And it also has a direction (the same as the displacement).
In this problem:
- The object travels a distance of
d = 105 m
In a time interval of
t = 22 s
So its speed is
- The displacement of the object is the same as the distance in this case, so still 105 m, covered in a time interval of 22 s; this means that the magnitude of the velocity is the same as the speed:
However, velocity is a vector quantity, so it also has a direction: and since the object has moved north, the direction of the velocity is north as well.