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 = 1\, s $ ?
Whether the particle is moving in positive $ x $ direction or negative $ x $ direction?

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

Answer 1

Answer: (B)

Given, position of the particle is
$x = 4 - 12 t + 3 t^2$
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 = 1\,s, v = 6\times 1 - 12 $
$ = -6\,m/s$
At $ t = 0\,s, x = 4\,m$
At $ t = 1\,s, x = 4 - (12 \times 1) + ( 3\times 1^2) $
$ = - 5\,m$
Hence, particle is moving along negative $x$ direction.