Answer to Question #344321 in Physics for Freddy101

Question

The distance S in metres covered by particle at time T, seconds is given by S= 2t²– 3t² – 5t, find:

(i) it's speed In the 5th second.

(ii) the distance covered in the 3rd second.

(iii) the acceleration at T=5 seconds.

Answer 1

$$s=2t^3-3t^2-5t$$

(i)

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

(ii)

$$d=s(5)-s(4)\\ =(2*5^3-3*5^2-5*5)\\-(2*4^3-3*4^2-4*5)=90\:\rm m$$

(iii)

$$a=12t-6\\ a(5)=12*5-6=54\:\rm m/s^2$$