Physics, asked by vaishnav7265, 1 year ago

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

Answers

Answered by talibbeiggcet
7
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
Answered by handgunmaine
3

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.

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