Thread: Average/Instantaneous rate of change questions.

1. Which function has a negative average rate of change on the interval 1<x<4?

a) f(x) = x^2 - x - 1
b) g(x) = 1.6x - 2
c) h(x) = -x^2 + 9
d j(x) = -3

2. For which value of x is the instantaneous rate of change of h(x) = 0.5x^2 + x - 2 closest to 0?

a) x= -1
b) x= -1
c) x= 0
d x= 1

3. Martin walks 5 m toward a motion sensor over the course of 10 s, at a constant speed. What would be the slope of the segment representing this walk on a distance versus time graph?

a) -2
b) -1/2
c) 1/2
d) 2

4. A student is walking in a straight line infront of a motion sensor. The sensor begins collecting data when the student is 6 m away. The student walks toward the sensor for 4 s at a rate of 1m/s. Then she walks away from the sensor for 8 s at a rate of 0.5 m/s. Which of the points is on the graph of the distance versus time?

a) (6, 0)
b) (8,2)
c) (10, 6)
d) (12,6)

5. Myra is riding a Ferris Wheel. Her height h(t), in metres above the ground at time t seconds, can be modelled by h(t) = 10sin(6(t-20)) + 10. At what time will Myra's car be at its greatest height?

a) t=20s
b) t=35s
c) t=40s
d) t=55s

6. At which point on the graph of f(x) = -x^2 -2x + 15 is the slope of the tangent 0?

a. (-2,15)
b. (-1, 16)
c. (0, 15)
d. (3,0)

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For the image below, (number 5 and 6)

for the first graph where it says, Which graph models walking directly away from a motion sensor at a constant rate?

It is letter C?

For number 6, is it D, cannot be determined because its staying still? Or could it be intervals 5<t<7 (b)?

Most are correct although I believe that #4 has two correct answers (initially, t= 0 and we are told she was 6 m from the sensor). 5 is correct assuming that "sin(x)" has x in degrees which is not standard when sine and cosine are used simply "as functions" rather than with angles. For 6, no, it is not "staying still". That is a graph of speed at time t, not position. A horizontal line means moving at that constant speed. And, as long as speed is constant, distance moved is just "speed times time", much like the area of a rectangle is "height time base"- which is why the distance moved is given by the area under the speed graph.

For h(t) = 10sin(6(t-20)) + 10, it is 20 degrees, it does not really matter does it?

For number #4, I know it is between these two answers (10, 6)
(12,6) since y= 6

For number 6, the answer has to be a then, interval 0<t<5?


By the way, is it ok if I post more questions in this thread to verify my answers? T.T

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In the graph below, it says A distance versus time graph is shown below. What is the instantaneous rate of change in the distance at t = 6s.

I did like this

First I found out the equation from 4 to 8 seconds, which gave me y=1/4x + 3.

Second, to find the instantaneous, I used values closer to 6s, -> 5.99 and 6.01

Third, I use the equation to calculate

y=1/4x + 3

y= 1/4(6.01) + 3
=4.5025

y= 1/4(5.99) + 3
= 4.4975

then I used Y2-Y1/X2-X1

=4.5025-4.4975/6.01-5.99
= 0.25 m/s?

Therefore, the instantaneous rate of change is 0.25 m/s.