The position of a particle moving along the x-axis is given by s(t)=2t^2+10.

Question

Use difference quotients to find the velocity v(t) and acceleration a(t).

Answer 1

"Use difference quotients" sounds like use the limit definition of the derivative. Otherwise, you would simply use the "power rule" to find the derivative of position to get the velocity function, and derivative of that for acceleration.

We'll use the difference quotients instead, which is more tedious, but we arrive at the same answers:

v(t) = s^'(t) = lim_h→0 (s(t+h) - s(t)) / h = lim_h→0 ((2t² - 4th +2h² + 10) - (2t² + 10)) / h = lim_h→0 (- 4th +2h²) / h = lim_h→0 (- 4t +2h) = -4t

a(t) = v^'(t) = lim_h→0 (v(t+h) - v(t)) / h = lim_h→0 (( - 4t - 4h) - (- 4t))/ h = lim_h→0 - 4 = - 4

Answer 2

s(t) = 2t^2 +10

v(t) = s'(t) = 4t

a(t) = v'(t) = 4

s'(t) = (s(t+h)-s(t))/h = (2(t+h)^2 +10 - 2t^2 -10)/h = (2t^2 + 4th+ 2h^2 - 2t^2)/h = (4th+2h^2)/h = 4t+2h

let h=0 to get s'(t) = 4t + 2(0) = 4t

do the same to get a(t) = v'(t)