Answer:
The velocity at that time would be [tex](-67.975\; \rm s)[/tex]. The velocity of this particle is continuously changing.
Explanation:
Differentiate the expression for position [tex]x[/tex] with respect to time [tex]t[/tex] to find an expression for velocity.
[tex]\begin{aligned}v(t) &= \frac{d}{dt}[x(t)] \\ &= \frac{d}{dt} \left[ 7.4 + 9.2\, t - 2.1\, t^{3}\right]\\ &= 9.2 - 6.3\, t^{2}\end{aligned}[/tex].
Hence, at [tex]t = 3.5\; \rm s[/tex], velocity would be [tex]v(3.5) = 9.2 - 6.3 \times (3.5)^{2} = -67.975\; \rm m[/tex].
Since velocity [tex]v(t)[/tex] changes with time [tex]t[/tex], the velocity of this particle is continuously changing.
