Question

if the position of an object changes with time t as x=t (t-2) then find the magnitude of net displacement of the object in the time interval t=0 when its velocity is zero.

Answer 1

Answer: 1



The position of the object is given as:

$x = t(t-2) \\ \\ x = t^2-2t$

At $t=0$, position is:

$x_0 = 0^2-2(0) = 0 \\ \\ \implies x_0 = 0$

Thus, initially, the object is at the Origin.

Now, we have to find the position when velocity becomes zero. 
Velocity is defined as the change of position per unit time.


That is, we can mathematically write velocity as:

$v = \frac{dx}{dt}$

We have position as a function of time. We can find velocity by differentiating position with respect to time.


$x=t^2-2t \\ \\ \\ \implies v = \frac{dx}{dt} = 2t - 2 \\ \\ \\ v = 2t-2$

Now, we have to find the time when velcocity becomes zero:

$2t-2=0 \\ \\ \\ \implies 2t=2 \\ \\ \\ \implies t = 1$

So, we know that velcocity becomes zero at t=1. Let us find the position at that time:

$x = t^2-2t \\ \\ \implies x_1 = (1)^2-2(1) \\ \\ \implies x_1 = 1-2 \\ \\ \implies x_1 = -1$

So, we have displacement as:

$Displacement = \text{Final Position - Initial Position} \\ \\ \implies Displacement = x_1 - x_0 \\ \\ \implies Displacement = -1 - 0 \\ \\ \implies Displacement = -1 \\ \\ \\ \implies \boxed{\text{Magnitude of Displacement } = 1}$