Question

The distance d of a particle moving in a straight line is given by d(t) = 2t3 + 5t – 2, where t is given in seconds and d is measured in meters. Find an expression for the instantaneous velocity v(t) of the particle at any given point in time. Question 1 options: 6t3 – 5 5t3 + 6 6t2 + 5 5t2 – 6

Answer 1

Answer:

(C)$6t^2+5$

Step-by-step explanation:

Given the distance, d(t) of a particle moving in a straight line at any time t is:

$d(t) = 2t^3 + 5t - 2, where t is given in seconds and d is measured in meters.$

To find an expression for the instantaneous velocity v(t) of the particle at any given point in time, we take the derivative of d(t).

$v(t)=\dfrac{d}{dt}\\\\v(t) =\dfrac{d}{dt}(2t^3 + 5t - 2) =3(2)t^{3-1}+5t^{1-1}\\\\v(t)=6t^2+5$

The correct option is C.