a train accelerates uniformly from 18 kilometre per hour to 36 km per hour in 10 seconds find the distance travelled
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33
u = 18 km/h = 18×5/18 = 5 m/s
v = 36 km/h = 36×5/18 = 10 m/s
t = 10 s
v = u+at
10 = 5 + 10a
10a = 10-5
10a = 5
a = 5/10
a = 0.5 m/s²
s = ut + 1/2at²
= 5×10+1/2×0.5×10²
= 50+0.25×100
= 50+25
= 75 m
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v = 36 km/h = 36×5/18 = 10 m/s
t = 10 s
v = u+at
10 = 5 + 10a
10a = 10-5
10a = 5
a = 5/10
a = 0.5 m/s²
s = ut + 1/2at²
= 5×10+1/2×0.5×10²
= 50+0.25×100
= 50+25
= 75 m
Hope it helps you ☺
Please mark as Brainliest ☺
Answered by
6
18 Km/h to 36 km/h in 10 seconds
V1=18Km/h,V2=36Km/h and Time=10 seconds
So we need to change it into proper units i.e. seconds..
Then,
18km/h × 1/60 × 1/60 =18/3600=0.005 km/sec
36km/h×1/60×1/60=>36/3600
=>1/100=0.01 km/sec
So,average velocity = V1+V2/2
0.01+0.005/2=0.015/2=0.0075
So Distance=Average velocity × Time =>
0.0075×10=0.075 km
So the Distance = 0.075 km or 75 metres.
V1=18Km/h,V2=36Km/h and Time=10 seconds
So we need to change it into proper units i.e. seconds..
Then,
18km/h × 1/60 × 1/60 =18/3600=0.005 km/sec
36km/h×1/60×1/60=>36/3600
=>1/100=0.01 km/sec
So,average velocity = V1+V2/2
0.01+0.005/2=0.015/2=0.0075
So Distance=Average velocity × Time =>
0.0075×10=0.075 km
So the Distance = 0.075 km or 75 metres.
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