Answer to Question #301332 in Physics for ezekiel

Question

The distance S in meters covered by a particle at time t seconds is given by

S=2t³-3t²-5t.Find

i) It's speed in the 5th second

ii) It's distance in the 3rd second

iii) The time the particle is momentarily at rest

iv) The acceleration at t=5s?

Answer 1

Given:

$S=2t³-3t²-5t$

i) It's speed in the 5th second

$v=S'=6t^2-6t-5\\ v(5)=6*5^2-6*5-5=115\:\rm m/s$

ii) It's distance in the 3rd second

$S(3)=2*3^3-3*3^2-5*3=12\:\rm m$

iii) The time the particle is momentarily at rest

$v=6t^2-6t-5=0$

$t=1.54\:\rm s$

iv) The acceleration at t=5s?

$a=v'=12t-6\\ a(5)=12*5-6=54\:\rm m/s^2$