physical applications q ! (1 Viewer)

[sammy_G]

somebody help me with this q PLEASE (actuali 3 small q's)

the acceleration of a particle P is given by the equation

a = 8x(x^2 + 1)

where x is the displacement of P from the origin in metres after t seconds, with movement being in a straight line. Initially, the particle is projected from the origin with a velocity of 2ms^-1 in the positive direction.

i) show that the velocity of the particle can be expressed as v=2(x^2 + 1)

ii) hence, show that the equation describing the displacement of the particle at time t is given by x = tan 2t

iii) Determine the velocity of the particle after pi/8 seconds

 

[onebytwo]

for the first one, let d/dx(1/2(v^2)) = a = 8x(x^2+1)
integrate a to get the value of 1/2 (v^2), then simplify to get v by itself

 

[hyparzero]

v=2(x² + 1)

=> dx/dt = 2(x² + 1)

rearrange and integrate and use when t=0 ; x = 0 etc

(integration is the inverse Tan forumla)

Hence

x = tan 2t

 

[followme]

a) a=8x(x<SUP>2</SUP>+1)
v<SUP>2</SUP>=2∫a dx = 2∫4*2x(x<SUP>2</SUP>+1) dx
move 4 out = 8[1/2(x<SUP>2</SUP>+1)<SUP>2</SUP>]+C
v<SUP>2</SUP>=4(x<SUP>2</SUP>+1)<SUP>2</SUP> + C (v=2 , x=0 therefore C=0)

so v=2(x<SUP>2</SUP>+1)

c) x=tan2t
v=2sec<SUP>2</SUP> 2t
t=pi/8 v=4 m/s or

x=tan2(pi/8) = 1
v = 2 (1+1) =4