a particle travel along a parabolic shape y=10+0.4x^2 with a constant speed of 6m/s at X=3m find components of velocity and acceleration
what direction is it going at x = 3 ?
it is tangent to the curve there
so what is the slope at x = 3?
dy/dx = 0.8 x = 2.4
Vx^2 + Vy^2 = 6^2 = 36
Vy/Vx = 2.4
solve for Vx and Vy
now for acceleration
there is no change in speed, only in direction
Vx^2 + Vy^2 = 36
Vy/Vx = 0.8 x so Vy = 0.8 x Vx
dVy/dt = 0.8 (x dVx/dt + Vx dx/dt) = 0.8 (x dVx/dt + Vx^2)
but 2 Vx dVx/dt + 2 Vy dVy/dt = 0 so dVy/ dt = -(Vx/Vy)dVx/dt
so
-(Vx/Vy) dVx/dt = 0.8(x dVx/dt + Vx^2)
you know x = 3 and Vx and Vy from the first part so solve fo dVx/dt, the x component of acceleration. Go back and get d Vy/dt