kinematics question..............?

Question

the velocity of a particle moving in the +ve direction of x-axis varies as v = 5x^1/2 .assuming that at t=0,v=0.find the acceleration of the particle at x=1 metre.......??


a particle moves in the direction of east for 25 secs with a velocity of 15 m/s .then,it moves towards north for 8 secs with a velocity of 5 m/s.the magnitude of average velocity will be...............??

Answer 1

v(x) = dx/dt = 5 sqrt (x)

dx / sqrt (x) = 5dt

2 sqrt(x) = 5t + C

because x=0 when t=0, C=0

so x = (5/2 t)^2 = 25/4 t^2
v = dx/dt = 25/2 t
a = dv/dt = 25/2

So acceleration is constant: 25/2.

Other question:
Displacement = 25*15 m east + 8*5 m north

multiply out and then use the pythagorean theorem to get the displacement magnitude:

d = sqrt (x^2 + y^2)

avarage speed = d/t

direction = arctan (y/x)

Edit--careful there eyes.
a = dv/dt (what you said), not dv/dx (what you did)

Answer 2

Not clear what you mean by "+ve" But if ve is a velocity variable along x, then dv/dt = a = 5/2 x^(-1/2) is the acceleration. And that is independent of t and v; so the fact that both are zero is inconsequential. Just let x = 1 in the equation for a and you have your acceleration.

The total time traveled is T = te + tn = 25 + 8 = 33 sec. The total distance traveled in that time is S = se + sn = ve*te + vn*tn = 15*25 + 5*8 as vectors. Thus S = sqrt(se^2 + sn^2) as scalars; where se = 15*25 and sn = 5*8. The average velocity V = S/T; you have the numbers, you can do the math.

Answer 3

v=5x^1/2
dx/dt=5x^1/2
dx/x^1/2=5dt
2x^1/2=5t+c
when t=0,v=0,x=0,
c=0, when x=1,=5,a=25/2m/s^2
2x^1/2=5t
a=dv/dt=5/2 v x^-1/2
2.towardseastx=15x25=375m
towards north,y =5x8=40m
distance between starting point and final position=sqrt(375^2+1600)=377.13m
average velocity=377.13/33=11.43m/saNs