Motion Under Variable Acceleration
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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.
1, 2, 3 are correct
2, 3 are correct
all of them are correct
1 and 2 are correct
A particle located at x = 0, at time t = 0 starts moving along the positive x direction with a velocity v that varies as v = α √x.The displacement of the particle varies with time as
t
t2
t1/2
t3
What if I wanted the displacement and distance of a body between times t = 0 and time t = 5 for V(t)=4−t2
None of these
displacement= -21.666, distance = 32.333
displacement= 32.333, distance = -21.666
displacement= 21.666, distance = 32.333
A particle moves according to the equation dvdt = α - β v , where α and β are constants. Find the velocity as a funtion of time. Assume body starts from rest.