Question

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

Answer 1

Answer:

see

Step-by-step explanation:

Final Answer : 50m/s^2 .

We know, that

velocity along x is constant so,

a(x) = 0

Now,

1)

\begin{gathered}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 \\\end{gathered}y=−x2+2x+2dtdy=−2xdtdx+2dtdx+0=>v(y)=−2x×5+2×5=>v(y)=−10x+10

Now,

\begin{gathered}a(y) = \frac{dv(y)}{dt} \\ = > - 10 \frac{dx}{dt} \\ = > - 10 \times 5 \\ = > - 50m \div {s}^{2}\end{gathered}a(y)=dtdv(y)=>−10dtdx=>−10×5=>−50m÷s2

Therefore, magnitude of acceleration is 50m/s^2.

Answer 2

Step-by-step explanation:

Final Answer : 50m/s^2 .

We know, that

velocity along x is constant so,

a(x) = 0

Now,

1)

\begin{gathered}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 \\\end{gathered}

y=−x

2

+2x+2

dt

dy

=−2x

dt

dx

+2

dt

dx

+0

=>v(y)=−2x×5+2×5

=>v(y)=−10x+10

Now,

\begin{gathered}a(y) = \frac{dv(y)}{dt} \\ = > - 10 \frac{dx}{dt} \\ = > - 10 \times 5 \\ = > - 50m \div {s}^{2}\end{gathered}

a(y)=

dt

dv(y)

=>−10

dt

dx

=>−10×5

=>−50m÷s

2

Therefore, magnitude of acceleration is 50m/s^2.