1)If the velocity v of a particle moving along a straight line decreases linearly with its displacement from 20m/s to a value approaching zero at s=30m,determine the acceleration of the particle when s=15m.
(Can u please solve this problem without the application of limits?)
2)A particle travels in a straight line such that for a short time 2s≤t≤6s,its motion is described by v=(4a)m/s,where a is m/s2.If v=6m/s when t=2s,determine the acceleration of the particle when t=3s.
1057202=2 x a x 30
a=-203 m/s2 negative since the velocity is decreasing...
i still dont understand....if the velocity is linearly decreasing then the acc is constant throughout...so i guess -203 m/s2 is the answer...
in the second question....v=4a where a is acceleration is dimensionally incorrect...
71)
Ketan: Here acc is not constant w.r.t time.
By some integration , i found out that v= 20 when x =15. and
again after some calcu.
i got acc = -40/3 ....
(Avik, what's the ans))
I may be wrong..
158ketan is correct...the correct answr is -203
@ketan: why is acceleration dimensionally wrong in the second problem??......
i have another question.....why is sn=u+at-12a dimensionally incorrect equation??
49[ sn ]=M0L1T0
[ u ]=M0L1T-1
[ at ]=M0L1T-1
[ 12a ]=M0L1T-2
Figuratively, in the right hand side you are adding a cow with a donkey and equating it with a horse (in the left hand side), savvy?
49No the second question isn't wrong.
The numerator can have a dimension of L2T-3, isn't it?
If the second question is wrong, then the questions like a=(8t2+16t)m/s2 would be wrong, isn't it?
1057@anik....what does sn denote?
if it denotes distance travelled in the nth second then the equation is correct except for the -1/2a part....
158ya...sn is distance travelled in nth second....
why is -12a wrong??
1057arre dimensionally...
sn dimension is [M0L1S-1] whereas -1/2a dimension is of acceleration which is [M0L1S-2]...
there ought to be a t attached to a....
49Actually the equation is not dimensionally incorrect. it is our perception and wrong method of writing which makes it seem so..
the eqn is: S=u+at-(a/2)x(1sec)