Question

if an object moves along a line so that it is at y = f(x)=7x^{2}-4x at time x (in seconds), find the instantaneous velocity function v = f(x), and find the velocity at times x = 1, 3, and 5 seconds (y is measured in feet).
v = f(x)=() ft/sec

Explanation:

Step1: Differentiate the position - function

The derivative of $y = f(x)=7x^{2}-4x$ using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$.
For the term $7x^{2}$, its derivative is $2\times7x^{2 - 1}=14x$. For the term $-4x$, its derivative is $-4$. So $f^{\prime}(x)=14x - 4$.

Step2: Find velocity at $x = 1$

Substitute $x = 1$ into $v=f^{\prime}(x)$.
$v=f^{\prime}(1)=14\times1-4=10$ ft/sec.

Step3: Find velocity at $x = 3$

Substitute $x = 3$ into $v=f^{\prime}(x)$.
$v=f^{\prime}(3)=14\times3-4=42 - 4=38$ ft/sec.

Step4: Find velocity at $x = 5$

Substitute $x = 5$ into $v=f^{\prime}(x)$.
$v=f^{\prime}(5)=14\times5-4=70 - 4=66$ ft/sec.

Answer:

$v = f^{\prime}(x)=14x - 4$; at $x = 1$, $v = 10$ ft/sec; at $x = 3$, $v = 38$ ft/sec; at $x = 5$, $v = 66$ ft/sec