HHApCalc5 2.1.004

Question

In a time of t seconds, a particle moves a distance of s meters from its starting point, where s = 5t² + 3.

(a) Find the average velocity between t = 1

 and t = 1 + h

 if each of the following is true.

(i) h=0.1

(ii) h=0.01

(iii) h=0.001

(b) Use your answers to part (a) to estimate the instantaneous velocity of the particle at time t=1.

Answer 1

(a) To solve part (a), we first need to define average velocity. In this case, it is the total distance traveled divided by the time it took to travel that particular distance. This makes sense intuitively. For example, if we travel 60 miles in 1 hour, our average velocity is 60 mph. This does not mean we were traveling exactly at 60 mph the entire time, but it is our average over the time interval. Let's apply that definition to this problem:

s(t) = 5t² + 3 (distance traveled as a function of time, t)

V_ave = (Distance traveled) / (Time interval)

Distance traveled = s(1 + h) - s(1)

Time interval = (1 + h) - (1) = h

s(1 + h) = 5*(1 + h)² + 3

s(1 + h) = 5*(1 + h)*(1 + h) + 3

s(1 + h) = 5*(1 + 2h +h²) + 3

s(1 + h) = 5 + 10h + 5h² + 3 = 5h² +10h + 8

s(1) = 5*(1)² + 3 = 8

V_ave = [s(1 + h) - s(1)] / h

V_ave = [(5h² + 10h + 8) - (8)] / h, (the 8's in the numerator cancel out, and a "5h" can be factored out)

V_ave = [5h*(h + 2)] / h, (cancel out an h on top and bottom)

V_ave = 5*(h + 2)

This is the key to our problem. Now we have a general expression for V_ave in terms of h, and parts (i), (ii), and (iii) are trivial.

(i) h = 0.1

V_ave = 5*(0.1+2) = 10.5

(ii) h = 0.01

V_ave = 5*(0.01+2) = 10.05

(iii) I will leave this as an exercise at this point.

(b) Even by inspection, we can see where we are headed as h gets smaller. We can intuitively estimate that the velocity is approaching 10. To get the instantaneous velocity, we simply use our general equation from part (a) and take the limit as h approaches 0. Our time increment keeps decreasing until it is infinitely small, at which point we have our instantaneous velocity.

V_inst = 5*(0 + 2) = 10