Question

The position of an atom moving inside a cathode ray tube is given by the function f(t) = t^3− 4t^2 + 3t where t is in seconds and f(t) is in meters. Find the instantaneous velocity of the atom at t = 2.5 seconds. A. 1.75 m/sec. B. 2.48 m/sec. C. 3.27 m/sec. D. 4.12 m/sec.

Answer 1

Instantaneous Velocity:

`\longrightarrow\:\rm v = (dt)/(dy)`

`\longrightarrow\:\rm v = (d)/(dy)( {t}^(3) - 4 {t}^(2) + 3t) \\`

`\longrightarrow\:\rm v = 3 {t}^(3 - 1) - 2.4 {t}^(2 - 1) + 3`

`\longrightarrow\:\rm v = 3 {t}^(2) - 8 t + 3`

`\longrightarrow\:\rm v = 3 {(2.5)}^(2) - 8 (2.5)+ 3`

`\longrightarrow\:\rm v = 3 * 6.25 - 20+ 3`

`\longrightarrow\:\rm v = 18.75 - 20+ 3`

`\longrightarrow\:\rm v = 18.75 - 20+ 3`

`\longrightarrow\:\bf v = 1.75 \: {ms}^( - 1)`

Answer 2

A

Explanation:

The position of an atom moving inside a cathode ray tube is given by the function:

`f(t)=t^3-4t^2+3t`

Where f(t) is in meters and t is in seconds.

And we want to determine its instantaneous velocity at t = 2.5 seconds.

The velocity function is the derivative of the position function. Thus, find the derivative of the function:

`f'(t)=3t^2-8t+3`

Then the instantaneous velocity at t = 2.5 will be:

`f'(2.5)=3(2.5)^2-8(2.5)+3=1.75\text{ m/sec}`

Our answer is A.