Question

If a particle's position is given by $x-4-12t+3t^2$ where t is in the seconds and $x$ in meters. What is its velocity at $t=1s$ ?
Whether the particle is moving in positive $x$ direction or negative $x$ direction?

• A. $-6m/s$ , $+x$ direction
• B. $-6m/s$ , $-x$ direction
• C. $6m/s$ , $+x$ direction
• D. $4m/s$ , $-x$ direction

Answer 1

Answer: (B)

Given, position of the particle is
$x=4-12t+3t^2$
Differentiating both sides with respect to time $(t)$,
we get
$\frac{dx}{dt}=0-12+3(2t)$
$=6t-12$
$\therefore v(t)=6t-12$
At $t=1s,v=6\times 1-12$
$=-6m/s$
At $t=0s,x=4m$
At $t=1s,x=4-(12\times 1)+(3\times 1^2)$
$=-5m$
Hence, particle is moving along negative $x$ direction.