54 kph = 15 m/s
the locomotive stops in 50 s
... (15 m/s) / (0.3 m/s^2) = 50 s
the average velocity is 7.5 m/s
... (15 m/s + 0 m/s) / 2 = 7.5 m/s
7.5 m/s * 50 s = 375 m
... 25 m from the traffic light
the locomotive stops in 50 s
... (15 m/s) / (0.3 m/s^2) = 50 s
the average velocity is 7.5 m/s
... (15 m/s + 0 m/s) / 2 = 7.5 m/s
7.5 m/s * 50 s = 375 m
... 25 m from the traffic light
distance l=400m
u=54kmph
a=-0.3m/s2
u=15mps
v=u+at
t=v-u/a
t=15s
avrage velocity=7.5m/s
distance s = 7.5/15=375m
total distance=400m-375m=25m
a = - 0.3 m/s2
∴ v = u + at
0 = 15 – 0.3 t
t = 15 / 0.3 = 50 sec
After 50 second, locomotive comes in rest permanently.
∴ the distance of the locomotive from traffic light =
400 – 375 = 25 m
1. Convert the velocity from km/hr to m/s:
To convert from km/hr to m/s, divide the velocity by 3.6.
Velocity (V) = 54 km/hr = 54/3.6 m/s = 15 m/s (rounded to two decimal places).
2. Convert the time from minutes to seconds:
Since the velocity is given in m/s, we need to convert time from minutes to seconds.
Time (t) = 1 minute = 1 × 60 seconds = 60 seconds.
3. Use the kinematic equation to calculate the displacement:
The kinematic equation relates displacement (s), initial velocity (u), time (t), and acceleration (a):
s = ut + (1/2)at^2
Given:
Initial velocity (u) = 15 m/s (because the brakes are applied)
Acceleration (a) = -0.3 m/s^2 (negative because it acts in the opposite direction of motion)
Time (t) = 60 seconds
Now substitute the values into the equation:
s = (15 × 60) + (1/2)(-0.3)(60^2)
s = 900 - (0.3 × 1800)
s = 900 - 540
s = 360 meters
Therefore, the position of the locomotive relative to the traffic light 1 minute after applying the brakes is 360 meters.