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

The velocity of the object as a function of time is v(t) = 4.8 + 12.8t, and the acceleration of the object as a function of time is a constant value of a(t) = 12.8 m/s².

Step-by-step explanation:

The student's provided position equation seems to have a typo and should likely be y(t) = 3.0 + 4.8t + 6.4t². The first step in determining the velocity of the object as a function of time is to take the first derivative of the position function with respect to time. In this case, the velocity v(t) will be the derivative of y(t), which is:

v(t) = d(3.0 + 4.8t + 6.4t²)/dt

= 0 + 4.8 + 2 × 6.4t

= 4.8 + 12.8t.

The acceleration of the object as a function of time is found by taking the derivative of the velocity function, which is:

a(t) = dv(t)/dt

= d(4.8 + 12.8t)/dt

= 0 + 12.8

= 12.8 m/s².

Acceleration is a constant value here, indicating uniform acceleration in the y direction.

Answer 2

Step-by-step 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`