Question

The displacement x of a particle moving in one dimension, under the action of force of a constant force is related to time t by the equation $t=\sqrt{x}+3$ where x is in meter and t is in second. Calculate the velocity at any time t.

Answer 1

Derivative of displacement with respect to time is velocity
$v=\dfrac{dx}{dt}$

Complete Step by step solution:
At time t displacement is $t=\sqrt{x}+3$.
In order to get the displacement x as a function of t we need to simplify the above equation.
$t=\sqrt{x}+3$
$t-3=\sqrt{x}$---- (1)
On squaring both sides of equation (1) we get,
${{(t-3)}^{2}}=x$
Now let us simplify the term by using the algebraic formula for the square of sum of terms.
$x={{t}^{2}}-6t+9$---------- (a)
As we know that the velocity v is the derivative of displacement of x with respect to time t.
So differentiating equation (a)
$v=\dfrac{dx}{dt}$
$\dfrac{dx}{dt}=\dfrac{d({{t}^{2}}-6t+9)}{dt}$
$\dfrac{dx}{dt}=2t-6$
$v=2t-6$--------- (2)
Hence the velocity at any time is $v=2t-6$.
Taking the case if at any time t= 3 putting the value in equation (2) we get
$v=2(3)-6$
$v=0$

Note:
In this derivation problem, we converted the displacement x from the time equation in question to the function of t, so that the displacement is easily distinguished by the function t. We developed a velocity equation for further derivation which can be used to find the velocity of any time t. A displacement is a vector whose length is the shortest distance from a point P undergoing motion from the original to the final position. The velocity of an object, with respect to a frame of reference, is the rate of change of its location and is a function of time.
Velocity is a derivative of displacement. Acceleration is a derivative of velocity. Acceleration derivatives are jerks.