A body moves along a line according to s(t)= t2 −9t2+24t. Find the positions when it is momentarily at rest…

Question

A body moves along a line according to s(t)= t2 −9t2+24t. Find the positions when it is momentarily at rest and when it's acceleration is zero

ScanSolve AI Tutor · Accepted Answer

Okay, Abbie, let's tackle this motion problem. I'll assume the position function is $s(t) = t^3 - 9t^2 + 24t$ , as $t^2 - 9t^2 + 24t$ would simplify to $-8t^2 + 24t$ , which has a constant acceleration (meaning acceleration is never zero for $t>0$ ).

Given the position function:

s(t) = t^3 - 9t^2 + 24t

1. Find the positions when the body is momentarily at rest.

Find the velocity function $v(t)$ by differentiating $s(t)$ with respect to $t$ .
$v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 9t^2 + 24t)$ $v(t) = 3t^2 - 18t + 24$
Set the velocity $v(t)$ to zero to find the times when the body is momentarily at rest.
$3t^2 - 18t + 24 = 0$
Divide the entire equation by 3:
$t^2 - 6t + 8 = 0$
Factor the quadratic equation:
$(t-2)(t-4) = 0$
This gives two times when the body is at rest:
$t = 2 \quad \text{or} \quad t = 4$
Substitute these times back into the position function $s(t)$ to find the positions.
For $t=2$ :
$s(2) = (2)^3 - 9(2)^2 + 24(2)$ $s(2) = 8 - 9(4) + 48$ $s(2) = 8 - 36 + 48$ $s(2) = 20$
For $t=4$ :
$s(4) = (4)^3 - 9(4)^2 + 24(4)$ $s(4) = 64 - 9(16) + 96$ $s(4) = 64 - 144 + 96$ $s(4) = 16$
The positions when the body is momentarily at rest are $\boxed{\textbf{20 units and 16 units}}$ .

2. Find the position when its acceleration is zero.
Find the acceleration function $a(t)$ by differentiating the velocity function $v(t)$ with respect to $t$ .
$a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 18t + 24)$ $a(t) = 6t - 18$
Set the acceleration $a(t)$ to zero to find the time when the acceleration is zero.
$6t - 18 = 0$ $6t = 18$ $t = 3$
Substitute this time back into the position function $s(t)$ to find the position.
For $t=3$ :
$s(3) = (3)^3 - 9(3)^2 + 24(3)$ $s(3) = 27 - 9(9) + 72$ $s(3) = 27 - 81 + 72$ $s(3) = 18$
The position when the acceleration is zero is $\boxed{\textbf{18 units}}$ .

Point	t	y
x-intercept	0	0
Maximum	1.5	18
x-intercept	3	0

A body moves along a line according to s(t)= t2 −9t2+24t. Find the positions when it is momentarily at rest and when it's acceleration is zero

Answer

Step-by-step solution

Find the velocity function $v(t)$ by differentiating $s(t)$ with respect to $t$ .

Set the velocity $v(t)$ to zero to find the times when the body is momentarily at rest.

Substitute these times back into the position function $s(t)$ to find the positions.

Find the acceleration function $a(t)$ by differentiating the velocity function $v(t)$ with respect to $t$ .

Set the acceleration $a(t)$ to zero to find the time when the acceleration is zero.

Substitute this time back into the position function $s(t)$ to find the position.

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A body moves along a line according to s(t)= t2 −9t2+24t. Find the positions when it is momentarily at rest and when it's acceleration is zero

Answer

Step-by-step solution

Find the velocity function v(t)v(t)v(t) by differentiating s(t)s(t)s(t) with respect to ttt.

Set the velocity v(t)v(t)v(t) to zero to find the times when the body is momentarily at rest.

Substitute these times back into the position function s(t)s(t)s(t) to find the positions.

Find the acceleration function a(t)a(t)a(t) by differentiating the velocity function v(t)v(t)v(t) with respect to ttt.

Set the acceleration a(t)a(t)a(t) to zero to find the time when the acceleration is zero.

Substitute this time back into the position function s(t)s(t)s(t) to find the position.

Try one yourself

Still stuck on this one?

More Physics Questions

Find the velocity function $v(t)$ by differentiating $s(t)$ with respect to $t$ .

Set the velocity $v(t)$ to zero to find the times when the body is momentarily at rest.

Substitute these times back into the position function $s(t)$ to find the positions.

Find the acceleration function $a(t)$ by differentiating the velocity function $v(t)$ with respect to $t$ .

Set the acceleration $a(t)$ to zero to find the time when the acceleration is zero.

Substitute this time back into the position function $s(t)$ to find the position.