Derivatives: Velocity + Acceleration

**break** · October 25th 2009, 01:58 PM

The equation of motion of a particle, where s is in meters and t is in seconds, is given by the equation below.

$s = 5t^3 - 4t$

i know that

$v(t) = 15t^2 - 4$
and
$a(t) = 30t$

The question is "Find the acceleration when the velocity is 0. (Round the answer to one decimal place.)"

I feel dumb asking this cause it seems like an easy problem, but i have no clue what the answer is.

**adkinsjr** · October 25th 2009, 02:00 PM

Originally Posted by break

The equation of motion of a particle, where s is in meters and t is in seconds, is given by the equation below.

$s = 5t^3 - 4t$

i know that

$v(t) = 15t^2 - 4$
and
$a(t) = 30t$

The question is "Find the acceleration when the velocity is 0. (Round the answer to one decimal place.)"

I feel dumb asking this cause it seems like an easy problem, but i have no clue what the answer is.

Just set the velocity equal to zero and solve the quadratic for t.

$15t^2-4=0$

Just solve that equation for t, and then plug that into $a(t)$ to obtain the acceleration.

$15t^2=4$

t= $\frac{2}{\sqrt{15}}$

It's conventional to rationalize the denominator. So t could be written as $\frac{2\sqrt{15}}{15}$

$a(\frac{2\sqrt{15}}{15})=30(\frac{2\sqrt{15}}{15}) =4\sqrt{15}$

Thread: Derivatives: Velocity + Acceleration

LinkBack

Thread Tools

Display

Derivatives: Velocity + Acceleration

Similar Math Help Forum Discussions

velocity and acceleration

[SOLVED] Velocity and Acceleration--Derivatives

acceleration/velocity

velocity/acceleration

Velocity and acceleration..

Search Tags