Answer:
The answer is 150 meters.
Explanation:
We know that,
u=0m/s
a=2m/s2
t=10s
for case 1.
Therefore, using the second equation of motion,
s=ut +1/2at2
s=1/2*2*100
=100metres.
Case 2.
v=0m/s because the train is deccelerating.
a=-1m/s
to find u,
v=u + at
0=u + -1*10
Therefore u=10m/s
using second equating of motion,
s=10*10 +1/2*-1*100
=50m
Therefore,
100+50=150metres.
Thanks.
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Answers & Comments
we know that s1=ut+1/2at²
since u=0
s1=1/2(2)(10)(10)
s1=100m and also v=u+at
v=20m/s
now let s2 be the distance travelled during its declaration ...a=-1,u=20,t=10
now s2 = 200-50=150
total distance travelled =100+150=250m
I hope this will help u ;)
Answer:
The answer is 150 meters.
Explanation:
We know that,
u=0m/s
a=2m/s2
t=10s
for case 1.
Therefore, using the second equation of motion,
s=ut +1/2at2
s=1/2*2*100
=100metres.
Case 2.
v=0m/s because the train is deccelerating.
a=-1m/s
t=10s
to find u,
v=u + at
0=u + -1*10
Therefore u=10m/s
using second equating of motion,
s=10*10 +1/2*-1*100
=50m
Therefore,
100+50=150metres.
Thanks.