Question

The distance x of a particle moving in one dimension under the action of a constant force is related to time t by the equation equals to under root x + 3 where x in metres and 3 seconds find the displacement of the particle when its velocity is zero

Answer 1

Hey there !

Solution :

Given :

Time ( t ) = √x + 3

Transposing 3 to the Left side we get,

√x = t - 3

We need the value of ' x '. So We must make it a rational number. Hence Squaring on both sides we get,

x = ( t - 3 )²

=> x = t² - 6t + 9   -----( Equation 1 )

Now to get the velocity we need to differentiate ' x ' with respect to Time.

=> Velocity = $\frac{d}{dt} ( t^2 - 6t + 9 )$

=> Velocity = $\frac{d}{dt} ( t^2 ) - 6 * \frac{d}{dt} ( t ) + \frac{d}{dx} ( 9 )$

=> Velocity = $2t - 6 * 1 + 0$

=> Velocity = 2t - 6.

Simplify the above equation we get,

= 2t - 6 = 0

= 2t = 6

=> t = 6 / 2 

=> t = 3 

Hence time is 3, So x will be equal to:

x = 3 - 3 = 0

Hence there will be zero ( 0 ) displacement.

Hope my answer helped :-)

Answer 2

The distance x of a particle moving in one dimension under the action of constant force is related to the time t by equation t=√x+3. Find displacement of a particle when its velocity is zero

Given Equation,

$t = \sqrt{x}+3\\\\\sqrt{x} = t - 3$

$x = (t-3)^2\\\\x = t^2 - 6t + 9$

Velocity = dx / dt.

Diffrentiate the equation w.r.t t:

$\dfrac{dx}{dt}=\dfrac{d(t^2 - 6t + 9)}{dt} \\\\v = 2t - 6\\\\0 = 2t - 6\\2t = 6\\t = 3$

When t = 3,:

$t = \sqrt x + 3\\\\3 = \sqrt x + 3\\\\x = 0$

Hence Displacement is zero.