Question

An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x +6.4x ,What is the velocity of the object as a function of time?
What is the acceleration of the object as a function of time?

Answer 1

Answer:

Explanation:

Instant Velocity and Acceleration

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

$v(x)=y'(x)$

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

$a(x)=v'(x)=y''(x)$

We are given the function for y

$y(x)=3.0+4.8x +6.4x^2$

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

$v(x)=y'(x)=4.8+12.8x$

And the acceleration is $a(x)=v'(x)=12.8$