Calculus Velocity

Question

The position of an object moving along a line is given by the function s(t)=−14t2+140t. Find the average velocity of the object over the following intervals.

​(a)​ [1, 55​]

​(b) ​[1, 44​]

​(c)​ [1, 33​]

​(d) ​[1, 1p+​h] where h>0 is any real number.

Answer 1

Average velocity is the distance traveled in the time interval divided by the time interval. The bracketed intervals are time points.

I believe part d is supposed to be the bracket {1, 1+h} in the limit h tends to zero....This is the instantaneous velocity or speed. Velocity is a vector having speed and direction.

S(55) = -14x(55)² +140x(55)

S(1) = -14x(1)² + 140x(1)

We want S(55) - s(1) divided by 55-1 as the average speed in this interval

For part d we can use differential calculus or look at (1 +h) ² = 1 +2h + h² on subtracting 1 from this we get 2h +h² and dividing by h yields 2 + h; this is the general term for any positive h greater than zero.... then in the limit h tends to zero we have the instantaneous speed term as

-14 x (2) + 140 and around the t=1 interval ...

around the t=p interval we would be looking at [p+h, p] and we see an average velocity of

-14(2p+h) + 140 and an instantaneous velocity letting h tend to zero of -14x2xp +140

This is the time rate of change for S(t) evaluated at t=p.

the time rate of change of y(t) = mt +b is the slope of the line y(t) being m

The time rate of change of the general quadratic in vertex form a(t+h)² + K is 2a(t+h) becomes 2ap fot t=p and h tends to zero.

Answer 2

The average velocity is simply [s(t₂) - s(t₁)]/(t₂ - t₁) and you need to plug in the values given.

Part d is preparation for the notion of what happens when h gets "little"!

I am going to assume that the "p" in the question is a typographical error so that the answer in part d is

14(2+h)+140.