A particle travels along the path y24x with a constant speed

A particle travels along the path y^2=4x with a constant speed of v = 5 m/s .

Determine the x and y components of the particle\'s velocity when the particle is at x = 6 m .

Determine the x and y components of the particle\'s acceleration when the particle is at x = 6 m

Solution

Let derivative w.r.t. time be denoted by : \'

y^2=4x

Taking derivative w.r.t. t we get:

2yy\'=4x\'

x\'=yy\'/2

Speed =5 so

sqrt(x\'^2+y\'^2)=5

x\'^2+y\'^2=25

y\'^2y^2/4+y\'^2=25

y\'^2(1+y^2/4)=25

y\'^2(1+x)=25

y\'^2=25/7

y\'=5/sqrt(7)

x\'=yy\'/2

y=sqrt(4x}=2\\sqrt{x}=2 sqrt{6}

x\'=(5/sqrt(7))(sqrt(6))=5sqrt(6/7)

x\'=5sqrt(6/7),y\'=5/sqrt(7)

x\'=yy\'/2 , x\'^2+y\'^2=25

So,

(yy\'/2)^2+y\'^2=25

y\'^2(y^2/4+1)=25

Differentiating w.r.t. t gives:

2y\'y\'\'(y^2/4+1)+y\'^2(yy\')/2=0

2y\'y\'\'(x+1)+y\'^2(yy\')/2=0

y\'=5/sqrt(7),x=6

2y\'\'(x+1)+y\'^2y/2=0

y\'\'=-(y\'^2y)/(4(x+1))

y\'\'=-(25/7* 2sqrt(6))/(4*7)=-25sqrt(6)/98

x\'=yy\'/2

Taking derivative gives:

x\'\'=yy\'\'/2+y\'^2/2

x\'\'=-25sqrt(6)*2sqrt(6)/98+25/14

x\'\'=-25*6*2/98+25/14

x\'\'=25(-6/49+1/14)

x\'\'=25(-12+7)/98

x\'\'=-125/98,y\'\'=-25sqrt(6)/98