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

Answer:

Instantaneous Velocity:

$\longrightarrow\:\rm v = \dfrac{dt}{dy}$

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

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

$\longrightarrow\:\rm v = 3 {t}^{2} - 8 t + 3$

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

$\longrightarrow\:\rm v = 3 \times 6.25 - 20+ 3$

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

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

$\longrightarrow\:\bf v = 1.75 \: {ms}^{ - 1}$

Answer 2

Answer:

A

Step-by-step 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.