kinematics problems?

Question

1. A particle moves in x-y plane with constant acceleration 'a' directed along the negative y axis the equation of motion has the form 'y = px - qx^2' ; where p and q are positive constants. find velocity of particle at origin.

2. In a car race car A takes 't' sec less than car B at the finish, and passes the finishing point with speed v m/s more than car B. Assuming that both car start from rest and travel with constant accelaration of 'p' and 'q' respectively. Show that
' v=(√ pq)t'

Answer 1

1. If y = px-qx^2, and the acceleration of the particle has no x acceleration, so d^2x/d^2t=0 and d^2y/d^2= - a

that means that dy/dt=vy(t)=-a*t +vy0
and y(t)=y0+vy0*t-.5*a*t^2
and dx/dt=vx(t)=vx
x(t)=x0+vx*t
for any given t
t=(x-x0)/vx
and
0=y0-y+vy0*t-.5*a*t^2

t=(-vy0 +/- sqrt(vy0^2+2*a*(y0-y))/
(2*(y0-y))
taking t out of the equations and doing a mess of algebra, you end up with an equation that has the form
0=c+b*y+d*y^2
where
c=4*y0^2(x-x0)-vx0*(vy0-vy0^2+2*a*y0)
b=vx0^2*a-8*x*y0+8*x0*y0
d=4*(x-x0)
apply the quadratic equation to get
y=(-b+/-sqrt(b^2-4*d*c))/(2*d)
multiply the two factors together and resolve all of the constants to the form y=px-qx^2

Then solve for vx and vy and set y=0 and x=0 to calculate the velocity at the origin.

2.
Car A equation of motion is
d=.5*p*l^2
where l is the time it crosses the finishing point
r=p*l is the speed

Car B equation of motion is
d=.5*q*w^2
where w is the time it crosses the finishing point
s=q*w is the speed

from the problem statement
w-l=t
and
r-s=v
also
the distance traveled is the same, so

.5*p*l^2=.5*q*w^2
or
p*l^2=q^w^2

since r-s=v and
r=p*l
s=q*w
then
v=p*l-q*w
combine all of the equations to eliminate l and w
0=(p-q)^2*t^2+
(p-q)*t*(v+q*t)-
q*t*(v+q*t)+
(1-p/q)*(v+q*t)^2
This will resolve to a quadratic expression for v, which can be expressed in the quadratic equation in terms of p,q and t

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