Question

A particle moves with velocity function v(t) = -t2 + 5t - 3, with v measured in feet per second and t measured in seconds. Find the acceleration of the particle at time t = 3 seconds

Answer 1

given V( t )= -t² +5t -3

differentiate both sides with respect to t

d( v) /dt =-2t + 5v

or d (v)/dt. when t =3 =-2(3) +5
acceleration. = -1
bcoz rate or change of velocity gives acceleration

Answer 2

The acceleration of the particle at time t = 3 seconds is $-1\ m/s^2$.

Explanation:

A particle moves with velocity function is given by :

$v(t)=-t^2+5t-3$

Where

v is in feet per second and t measured in seconds

The acceleration of an object is given by :

$a=\dfrac{dv}{dt}$

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

$a=-2t+5$

At t = 3 second

$a=-2(3)+5$

$a=-1\ m/s^2$

So, the acceleration of the particle at time t = 3 seconds is $-1\ m/s^2$.

Learn more,

Velocity and acceleration

https://brainly.in/question/10876646