Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 2 - Motion Along a Straight Line - Problems: 15b

Answer

At $t=1s$, the particle is moving along the negative $x$ direction. This is because the displacement is given by the equation $x=4-12t+3t^2$. After differentiating this equation we get velocity, $v = -12+6t$. At time $t=1s$, velocity $v$ is $-12+6\times(1) = -6ms^{-1}$. This negative velocity implies that the particle is moving in the negative direction of $x$.

Work Step by Step

Step 1: Find the velocity equation by differentiating the given displacement equation. $$x=4-12t+3t^2$$ $$\implies v = -12+6t$$ Step 2: Plug in the value of time for which you want to calculate the velocity. Here, we want to calculate the velocity of the particle at time $t=1s$. So, $$v=-12+6\times(1)=-6m/s$$ Step 3: Note the sign of the velocity. Here it is negative which means that the particle is moving along the $-x$ direction.
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