Question

A particle moves along the parabolic path y=2x - x2 +2 in such a way that the x component of velocity vector remains constant (5m/s). Find the magnitude of acceleration of the particle.

Answer 1

Final Answer : 50m/s^2 .
We know, that
velocity along x is constant so,
a(x) = 0

Now,
1)
$y = - {x}^{2} + 2x + 2 \\ \frac{dy}{dt} = - 2x \frac{dx}{dt} + 2 \frac{dx}{dt} + 0 \\ = > v(y) = - 2x \times 5 + 2 \times 5 \\ = > v(y) = - 10x + 10$

Now,
$a(y) = \frac{dv(y)}{dt} \\ = > - 10 \frac{dx}{dt} \\ = > - 10 \times 5 \\ = > - 50m \div {s}^{2}$
Therefore, magnitude of acceleration is 50m/s^2.

Answer 2

answer is in the above attachment