Question
The velocity of a particle moving along a straight line increases according to the linear law v = v0 + kx
where k is a constant . Then
1) the accelearation of the particle is k(v0 + kx)
2) the particle takes a time 1k loge (v1v0) to attain a velocity v1
3) velocity varies linearly with displacement with slope of velocity displacement curve equal to k
4) data is insufficient to arrive at a conclusion.
where k is a constant . Then
1) the accelearation of the particle is k(v0 + kx)
2) the particle takes a time 1k loge (v1v0) to attain a velocity v1
3) velocity varies linearly with displacement with slope of velocity displacement curve equal to k
4) data is insufficient to arrive at a conclusion.
Answer: Option A
:
A
Options (1), (2), (3) are correct.
Accelearation = dvdt =kv
⇒ a = k (v0 + kx)
Further
⇒ dvdt = kv
⇒ dvv = kdt
v1∫v0 dvv = k t∫0 ft
t = 1k loge v1v0
Since , v = v0 + kx . Hence slope of velocity displacement curve is dvdx = k
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:
A
Options (1), (2), (3) are correct.
Accelearation = dvdt =kv
⇒ a = k (v0 + kx)
Further
⇒ dvdt = kv
⇒ dvv = kdt
v1∫v0 dvv = k t∫0 ft
t = 1k loge v1v0
Since , v = v0 + kx . Hence slope of velocity displacement curve is dvdx = k
Was this answer helpful ?
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