Question

$v(t) = t^4 - 2t^3 - 4t$
what is the particles velocity $v(t)$ at $t = 1$
$v(1) = square$
what is the particles acceleration $a(t)$ at $t = 1$
$a(1) = square$
at $t = 1$, is the particle speeding up, slowing down, or neither?

Explanation:

Step1: Substitute t=1 into v(t)

$v(1) = 1^4 - 2(1)^3 - 4(1)$

Step2: Calculate v(1)

$v(1) = 1 - 2 - 4 = -5$

Step3: Find acceleration a(t) (derivative of v(t))

$a(t) = v'(t) = 4t^3 - 6t^2 - 4$

Step4: Substitute t=1 into a(t)

$a(1) = 4(1)^3 - 6(1)^2 - 4$

Step5: Calculate a(1)

$a(1) = 4 - 6 - 4 = -6$

Step6: Compare signs of v(1) and a(1)

Both $v(1)=-5$ and $a(1)=-6$ are negative (same sign), so the particle is speeding up.

Answer:

$v(1) = -5$
$a(1) = -6$
The particle is speeding up.