Question

question 3
suppose a particle moves along the ( x )-axis and that its position after ( t ) seconds given by the function ( s(t) ). suppose further that the derivative of ( s(t) ) is given by ( s(t) = 2t^2 - 4t + 3 ). what is the instantaneous velocity of the particle after 2 seconds?
( \bigcirc 4 )
( \bigcirc 3 )
( \bigcirc \frac{10}{3} )
( \bigcirc \frac{1}{6} )

Explanation:

Step1: Recall velocity and derivative relation

The instantaneous velocity \( v(t) \) of a particle with position function \( s(t) \) is given by the derivative \( s'(t) \), so \( v(t)=s'(t) = 2t^{2}-4t + 3 \).

Step2: Substitute \( t = 2 \) into \( v(t) \)

Substitute \( t = 2 \) into the velocity function:
\( v(2)=2(2)^{2}-4(2)+3 \)
First, calculate \( (2)^{2}=4 \), then \( 2\times4 = 8 \), \( 4\times2=8 \).
So \( v(2)=8 - 8+3 \)
Simplify the expression: \( 8 - 8 = 0 \), then \( 0 + 3=3 \).

Answer:

3 (corresponding to the option with value 3)