## Fundamentals of Physics Extended (10th Edition)

In this question, a particle's position has been said to be $x = 4-12t+3t^2$ at time $t$ ($t$ in seconds and $x$ in meters). That is, the position of the particle at time $t=0s$ is $4-12\times0 + 3\times0^2 = 4m$ and at time $t=1s$ is $4-12\times1+3\times1^2 = -5m$ from the origin of a coordinate axes (though it is not specified in the question, we can consider it for our understanding). Here, a negative sign just means the opposite direction. You can think positive x-direction as positive displacement whereas the negative x-direction as the negative displacement. By the definition of velocity, we understand that the rate of change of displacement is velocity. So, the derivative of the displacement with respect to time is the particle's velocity. Thus, $$v = \frac{dx}{dt} = -12 + 6t$$ So, the velocity of the particle at time $t=1s$ is, $$v_{t=1}= \bigg[\frac{dx}{dt}\bigg]_{t=1} = -12 + 6\times(1) = -6ms^{-1}$$ Here, the negative value of velocity denotes that the particle is moving along the negative direction of x.
Step 1: Differentiate the given displacement equation with respect to time $t$ because the time-rate change of displacement is velocity. That is, $$v = \frac{dx}{dt} = \frac{d(4-12t+3t^2)}{dt} = -12 + 6t$$ Step 2: To calculate the velocity of the particle at time $t$, plug the value of $t$ into the new equation that we got after differentiating the displacement equation. $$v_{t=1}= \bigg[\frac{dx}{dt}\bigg]_{t=1} = -12 + 6\times(1) = -6ms^{-1}.$$ Step 3: Do check the sign of the answer that you got. Here, the sign is negative. It means that the velocity of the particle at time $t=1s$ is along the negative direction of x.