A body travels 200 cm in the first 2s and 220 cm in the next 5sec Calculate the velocity of the body at the end of 7sec from t he start assuming uniform acceleration through out the journey
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velocity
=total displacement/total time
=200+220/2+5
=420/7
=60m/s
=total displacement/total time
=200+220/2+5
=420/7
=60m/s
ntiwari215:
but the answer given is 2.22 m/s
Answered by
1
=200+220/2+5
=420/7
=60m/s
but the answer given is 2.22 m/s
i think the question is incomplete
(200 + 220)cm = 420cm
t' = (2 + 5)s = 7s
If a is the uniform acceleration of the particle,
then x(t) = x(0) + v(0)t + ½ at^2
200 = 0 + u*2 + 0.5*a *4
or 100 = u + a ..(1)
Again, x(t') = x(0) + v(0)t' + ½ at^2
420 = 0 + u*7 + 0.5*a*49
60 = u + 3.5a .. (2)
-40 = 2.5a
or a = -16cm/s
u = 100 - (-16) = 116cm/s, t'' = 7s, v(t'') = ?
v(t'') = v(0) + at'' v(t'') = (116 - 16*7) = 4cm/s.
2m in 2s.
Then, 2.2m in 5s it decelerating. 2.22m/s.
(please help me to understand this solution)
t' = (2 + 5)s = 7s
If a is the uniform acceleration of the particle,
then x(t) = x(0) + v(0)t + ½ at^2
200 = 0 + u*2 + 0.5*a *4
or 100 = u + a ..(1)
Again, x(t') = x(0) + v(0)t' + ½ at^2
420 = 0 + u*7 + 0.5*a*49
60 = u + 3.5a .. (2)
-40 = 2.5a
or a = -16cm/s
u = 100 - (-16) = 116cm/s, t'' = 7s, v(t'') = ?
v(t'') = v(0) + at'' v(t'') = (116 - 16*7) = 4cm/s.
2m in 2s.
Then, 2.2m in 5s it decelerating. 2.22m/s.
(please help me to understand this solution)
this solution is given in merritnation.com
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