Question

Show that the velocity of the particle at time t is given by v(t)=70t(3t2+8)2 .

Answer 1

Question:

A particle moves along the x-axis so that its position at time t>0 is given by $x(t)=\frac{t^2 -9}{3t^2 +8}$

Show that the velocity of the particle at time t is given by $v(t) = \frac{70t}{(3t^2 + 8)^2}$

Answer:

$v(t) = \frac{70t}{(3t^2 + 8)^2}$

Step-by-step explanation:

Given

$x(t)=\frac{t^2 -9}{3t^2 +8}$

Required

Show that:

$v(t) = \frac{70t}{(3t^2 +8)^2}$

To calculate v(t), we use:

$v(t) = x'(t)$

So, we have:

$x(t)=\frac{t^2 -9}{3t^2 +8}$

Apply quotient rule

$x'(t) = \frac{vu' - uv'}{v^2}$

Where

$u =t^2 - 9$

$v=3t^2 + 8$

So:

$u =t^2 - 9$

$u' = 2t$

$v=3t^2 + 8$

$v' = 6t$

So, we have:

$x'(t) = \frac{vu' - uv'}{v^2}$

$x'(t) = \frac{(3t^2 + 8) * 2t - (t^2 - 9) * 6t}{(3t^2 + 8)^2}$

Expand the numerator

$x'(t) = \frac{6t^3 + 16t - 6t^3 + 54t}{(3t^2 + 8)^2}$

Collect like terms

$x'(t) = \frac{6t^3 - 6t^3+ 16t + 54t}{(3t^2 + 8)^2}$

$x'(t) = \frac{70t}{(3t^2 + 8)^2}$

Recall that:

$v(t) = x'(t)$

$v(t) = \frac{70t}{(3t^2 + 8)^2}$

Proved