Question

if an object moves along the y - axis (marked in feet) so that its position at time x (in seconds) is given by f(x)=192x - 16x^2, find the following. (a) the instantaneous velocity function v = f(x) (b) the velocity when x = 0 and x = 4 sec (c) the time(s) when v = 0 (a) v(x)=□

Explanation:

Step1: Differentiate the position - function

The position function is $f(x)=192x - 16x^{2}$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have:
$v(x)=f^{\prime}(x)=\frac{d}{dx}(192x-16x^{2})=\frac{d}{dx}(192x)-\frac{d}{dx}(16x^{2})$.
Since $\frac{d}{dx}(192x)=192$ and $\frac{d}{dx}(16x^{2}) = 2\times16x=32x$, then $v(x)=192 - 32x$.

Step2: Find the velocity at $x = 0$ and $x = 4$

When $x = 0$, substitute into $v(x)$:
$v(0)=192-32\times0=192$ feet/sec.
When $x = 4$, substitute into $v(x)$:
$v(4)=192-32\times4=192 - 128 = 64$ feet/sec.

Step3: Find the time when $v = 0$

Set $v(x)=0$, so $192 - 32x=0$.
Add $32x$ to both sides: $192=32x$.
Divide both sides by 32: $x = 6$ seconds.

Answer:

(A) $v(x)=192 - 32x$
(B) When $x = 0$, $v = 192$ feet/sec; when $x = 4$, $v = 64$ feet/sec
(C) $x = 6$ seconds