a train moving 108 km per hour is brought to rest in 20 seconds find the acceleration and distance covered in 20 seconds and also find the time when its speed was 36 km per hour
Answers
Answered by
0
time - 20 secs
speed-108km/h
acceleration- v-u/t
u=108*5/18=30m/s
v=0m/s
-> 0-30/20= -3/2 = -1.5m/s is the acceleration
distance= ut+1/2at^2
-> 108*20+1/2*-30*20*20
= 2160-6000
= 3840 metres
case2 -
speed-36kmph
distance-3840
time - ?
time=distance/speed
-> 106 seconds
speed-108km/h
acceleration- v-u/t
u=108*5/18=30m/s
v=0m/s
-> 0-30/20= -3/2 = -1.5m/s is the acceleration
distance= ut+1/2at^2
-> 108*20+1/2*-30*20*20
= 2160-6000
= 3840 metres
case2 -
speed-36kmph
distance-3840
time - ?
time=distance/speed
-> 106 seconds
Similar questions