Question

The position of an object at time t is given by s(t) = -9 - 3t. Find the instantaneous velocity at t = 8 by finding the derivative.

I absolutely have to use the "delta h format"

Answer 1

If you've taken calculus, the easiest way to solve this is to take the derivative of s(t):

2t-3. This gives you the rate that the particle is moving at for t>0. Thus the particle will be at a maximum/minimum whenever the derivative is 0 (or at the end points), and it will change direction when the rate it's moving (2t-3) goes from positive to negative or vice versa.

So at t=1.5 the particle has stopped. s(0) = 2, and at first the particle is moving left, until it gets to t=1.5 at which point it turns around and goes positive. Continues positive for all t then.

If you haven't taken Calculus it's a little bit more complicated, but not bad. In that case the easiest way is probably to figure out what it looks like.

because t^2-3t+2 = (t-2)(t-1), you know that the function is zero at t=1, 2. you also know it's an x^2 function, so you know it curves like a parabola (turns around once). Now try s(1.5)=4.25-4.5 = -.25

Thus it's negative between 1 and 2, and positive otherwise. So you can now see how the particle moves: starts out positive, moves negative toward s(1.5), then turns around and goes positive.

Answer 2

by finding the derivative get an expression of velocity then put t=8 into it.

Answer 3

Basically for this question all I need to do is find the derivative. I know that the derivative is -3 and that will be my answer. I just dont know how to get to that conclusion using the delta process (which is what my teacher wants to see)

Answer 4

I think I know how to do it now.

Answer 5

i'm trying to print delta from the equation editor but cant

Answer 6

lim h-->8 (f(x+h) - f(x))/h

Answer 7

thats it

Answer 8

using the difference quotient

Answer 9

what you are doing is finding the derivative from first principles

Answer 10

ok. thanks.