Question

A particle moves with velocity function v(t) = ^{2} " alt=" 3t^{2} " align="absmiddle" class="latex-formula"> − 4t + 1, with v measured in feet per second and t measured in seconds. Find the acceleration of the particle at time t = 3 seconds.
2 divided by 3 feet per second^2
7 feet per second^2
14 feet per second^2
21 feet per second^2

Answer 1

V'(t)=a(t)

deriviive of the velocity is the acceleration

easy peasy
dy/dx v(t)=6t-4
v'(t)=a(t)=6t-4

at=t=3
a(3)=6(3)-4
a(3)=18-4
a(3)=14 ft per second squared

3rd answer

Answer 2

Acceleration is the rate of change of velocity.
So take the derivative of the velocity which would be a =6t-4.  When t=3, a=14 ft/sec^2