Question:
A particle moves along the x-axis so that its position at time t>0 is given by [tex]x(t)=\frac{t^2 -9}{3t^2 +8}[/tex]
Show that the velocity of the particle at time t is given by [tex]v(t) = \frac{70t}{(3t^2 + 8)^2}[/tex]
Answer:
[tex]v(t) = \frac{70t}{(3t^2 + 8)^2}[/tex]
Step-by-step explanation:
Given
[tex]x(t)=\frac{t^2 -9}{3t^2 +8}[/tex]
Required
Show that:
[tex]v(t) = \frac{70t}{(3t^2 +8)^2}[/tex]
To calculate v(t), we use:
[tex]v(t) = x'(t)[/tex]
So, we have:
[tex]x(t)=\frac{t^2 -9}{3t^2 +8}[/tex]
Apply quotient rule
[tex]x'(t) = \frac{vu' - uv'}{v^2}[/tex]
Where
[tex]u =t^2 - 9[/tex]
[tex]v=3t^2 + 8[/tex]
So:
[tex]u =t^2 - 9[/tex]
[tex]u' = 2t[/tex]
[tex]v=3t^2 + 8[/tex]
[tex]v' = 6t[/tex]
So, we have:
[tex]x'(t) = \frac{vu' - uv'}{v^2}[/tex]
[tex]x'(t) = \frac{(3t^2 + 8) * 2t - (t^2 - 9) * 6t}{(3t^2 + 8)^2}[/tex]
Expand the numerator
[tex]x'(t) = \frac{6t^3 + 16t - 6t^3 + 54t}{(3t^2 + 8)^2}[/tex]
Collect like terms
[tex]x'(t) = \frac{6t^3 - 6t^3+ 16t + 54t}{(3t^2 + 8)^2}[/tex]
[tex]x'(t) = \frac{70t}{(3t^2 + 8)^2}[/tex]
Recall that:
[tex]v(t) = x'(t)[/tex]
[tex]v(t) = \frac{70t}{(3t^2 + 8)^2}[/tex]
Proved
