At a distance L equals to 400m from the traffic light, brakes are applied to a locomotive moving at a velociy VEquals to 54 km/hr. Determine the position of the locomotive relative to the traffic light l minute after application ofthe brakes if its acceleration is -0.3m/sec..
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
u = 54 x 5/18 = 15 m /s
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
given,
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
To determine the position of the locomotive relative to the traffic light, we need to calculate the distance traveled by the locomotive during the given time interval. Here are the steps to find the answer:
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.