Distance = 3600 m
Average Speed = 12 m/s
Time = 3600/12 = 300 sec
You cannot obtain the maximum speed from this information alone, for that you require some information regarding acceleration.
There is motion of a straight track between two stations.
Average speed: 12m/s
Distance between stations: 3600m.
Please help how could I find the max speed given these information? and it isn't like the average speed so it confused me
3 answers
Let us assume that there is a constant acceleration till the the middle of the stations, and from the middle point the vehicle decelerates at the same rate.
So, we can obtain the max speed, which would be at the middle point, by using equations of motion for half the journey.
s = 1800m
t = 150 sec (Since the journey is symmetrical)
s = (1/2)at^2
=> 1800 = (1/2)(a)(150)^2
=> 3600/22500 = a
=> a = 36/225
= 12/69 m/s^2
Now,
v = u + at
=> v = 0 + (12/69)*150
=> v = 1800/69
= 26 m/s
So, we can obtain the max speed, which would be at the middle point, by using equations of motion for half the journey.
s = 1800m
t = 150 sec (Since the journey is symmetrical)
s = (1/2)at^2
=> 1800 = (1/2)(a)(150)^2
=> 3600/22500 = a
=> a = 36/225
= 12/69 m/s^2
Now,
v = u + at
=> v = 0 + (12/69)*150
=> v = 1800/69
= 26 m/s
THANK U LOTS!
1 answer