The velocity of a particle moving along a straight line is given by v(t)=6t2+4t−5 cm/sec at time t seconds with initial position s(0)=3 cm. What is the position of the particle at t=2 seconds, in cm?

s(2) = 17 cm

Step-by-step explanation:

We are told that the velocity function is;

v(t) = 6t² + 4t − 5 cm/sec

Integral of velocity gives distance.

Thus;

s(t) = ∫v(t) = ∫6t² + 4t − 5

s(t) = 2t³ + 2t² - 5t + c

We are told that s(0)=3 cm

Thus;

s(0) = 2(0)³ + 2(0)² - 5(0) + c = 3

Thus; c = 3

Thus;

s(t) = 2t³ + 2t² - 5t + 3

At t = 2 secs

s(2) = 2(2)³ + 2(2)² - 5(2) + 3

s(2) = 17 cm

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb