Question

A particle moves along the positive branch of the curve y=x²/2 where x=t²/2, x and y measured in metres and t in second. At t=2s, the velocity of the projectile is

Answer 1

Answer:

2√5

Explanation:

x = t^2/2

y = x^2/2 = t^4 /8

At t = 2

Vx = dx/dt = t = 2

Vy = dy/dy = t^3 /2 = 4

Velocity is √(Vx^2 + Vy^2)

= 2√5

Answer 2

Given that :

A particle moves along the positive branch of the curve as follows :

$y=\dfrac{x^2}{2}$

Where

$x=\dfrac{t^2}{2}$

x and y measured in metres and t in second

To find :

Velocity of the projectile at t = 2s

Solution :

$y=\dfrac{t^4}{8}$

Velocity in x direction is :

$v_x=\dfrac{dx}{dt}\\\\v_x=\dfrac{d(t^2/2)}{dt}\\\\v_x=t$

Velocity in y direction is :

$v_y=\dfrac{dy}{dt}\\\\v_y=\dfrac{d(t^4/4)}{dt}\\\\v_y=\dfrac{t^3}{2}$

At t = 2 s,

$v_x=2\ m/s$

$v_y=\dfrac{2^3}{2}\\\\v_y=4\ m/s$

Net velocity :

$v=(2i+4j)\ m/s$

So, the velocity of the projectile is (2i+4j) m/s.

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Projectile motion

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