exercise 1.3
part a
1. the velocity of an object is given by ( v(t) = t(t - 4)^2 ), in seconds. at what times, in seconds, is the object at rest?
2. give a geometrical interpretation of the following expressions, if ( s ) is a position function:
a. ( \frac{s(9) - s(2)}{7} )
b. ( limlimits_{h o 0} \frac{s(6 + h) - s(6)}{h} )
c. ( limlimits_{h o 0} \frac{sqrt{4 + h} - 2}{h} )
3. give a geometrical interpretation of ( limlimits_{h o 0} \frac{s(6 + h) - s(6)}{h} ).
4. use the graph to answer each question.
(there is a graph of ( y = f(x) ) with points a, b, c, d, e)
a. between which two consecutive points is the average rate of change in the function the greatest?
b. is the average rate of change in the function between a and b greater than or less than the instantaneous rate of change at b?
c. sketch a tangent to the graph somewhere between d and e such that the slope of the tangent is the same as the average rate of change in the function between b and c.
5. what is wrong with the statement “the speed of the cheetah was 65 km/h north”?
6. is there anything wrong with the statement “a school bus had a velocity of 60 km/h for the morning run, which is why it was late arriving”?
part b
7. a construction worker drops a bolt while working on a high - rise building, 320 m above the ground. after ( t ) seconds, the bolt is at a height of ( s(t)=320 - 5t^{2},0leq tleq8 ), in metres, where ( s(t) ).
a. calculate the average velocity during the first, third, and eighth seconds.
b. calculate the average velocity for the interval ( 3leq tleq8 ).
c. calculate the velocity at ( t = 2 ).

Question

exercise 1.3
part a
1. the velocity of an object is given by ( v(t) = t(t - 4)^2 ), in seconds. at what times, in seconds, is the object at rest?
2. give a geometrical interpretation of the following expressions, if ( s ) is a position function:
   a. ( \frac{s(9) - s(2)}{7} )
   b. ( limlimits_{h 	o 0} \frac{s(6 + h) - s(6)}{h} )
   c. ( limlimits_{h 	o 0} \frac{sqrt{4 + h} - 2}{h} )
3. give a geometrical interpretation of ( limlimits_{h 	o 0} \frac{s(6 + h) - s(6)}{h} ).
4. use the graph to answer each question.
   (there is a graph of ( y = f(x) ) with points a, b, c, d, e)
   a. between which two consecutive points is the average rate of change in the function the greatest?
   b. is the average rate of change in the function between a and b greater than or less than the instantaneous rate of change at b?
   c. sketch a tangent to the graph somewhere between d and e such that the slope of the tangent is the same as the average rate of change in the function between b and c.
5. what is wrong with the statement “the speed of the cheetah was 65 km/h north”?
6. is there anything wrong with the statement “a school bus had a velocity of 60 km/h for the morning run, which is why it was late arriving”?
part b
7. a construction worker drops a bolt while working on a high - rise building, 320 m above the ground. after ( t ) seconds, the bolt is at a height of ( s(t)=320 - 5t^{2},0leq tleq8 ), in metres, where ( s(t) ).
   a. calculate the average velocity during the first, third, and eighth seconds.
   b. calculate the average velocity for the interval ( 3leq tleq8 ).
   c. calculate the velocity at ( t = 2 ).

Accepted Answer

### PART A
#### 1.
# Explanation:
## Step 1: Recall when velocity is zero
An object is at rest when its velocity $ v(t) = 0 $. Given $ v(t)=(t - 4)^2 $.
Set $ (t - 4)^2=0 $.
## Step 2: Solve for $ t $
Taking square roots on both sides, $ t - 4 = 0$, so $ t = 4 $.
# Answer:
The object is at rest when $ t = 4 $ seconds.

#### 2.
##### a.
# Explanation:
The expression $ \frac{s(9)-s(2)}{7} $ represents the average rate of change of the position function $ s(t) $ over the interval from $ t = 2 $ to $ t=9 $. The average rate of change of a function $ f(t) $ over the interval $[a,b]$ is given by $ \frac{f(b)-f(a)}{b - a} $, here $ a = 2 $, $ b=9 $ and $ b - a=7 $, so it's the average velocity (since velocity is the rate of change of position) from $ t = 2 $ to $ t = 9 $.
##### b.
# Explanation:
The limit $ \lim_{h
ightarrow0}\frac{s(6 + h)-s(6)}{h} $ represents the instantaneous rate of change of the position function $ s(t) $ at $ t = 6 $. By the definition of the derivative, $ f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h} $, so this limit is the instantaneous velocity (derivative of position function) at $ t = 6 $.

#### 3.
# Explanation:
The limit $ \lim_{h
ightarrow0}\frac{\sqrt{4+h}-2}{h} $ can be recognized as the definition of the derivative of the function $ f(x)=\sqrt{x} $ at $ x = 4 $. Using the derivative formula $ f^{\prime}(x)=\frac{1}{2\sqrt{x}} $, at $ x = 4 $, $ f^{\prime}(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4} $. We can also rationalize the numerator:
$$
\begin{align*}
\lim_{h
ightarrow0}\frac{\sqrt{4+h}-2}{h}&=\lim_{h
ightarrow0}\frac{(\sqrt{4+h}-2)(\sqrt{4+h}+2)}{h(\sqrt{4+h}+2)}\
&=\lim_{h
ightarrow0}\frac{(4 + h)-4}{h(\sqrt{4+h}+2)}\
&=\lim_{h
ightarrow0}\frac{h}{h(\sqrt{4+h}+2)}\
&=\lim_{h
ightarrow0}\frac{1}{\sqrt{4+h}+2}\
&=\frac{1}{\sqrt{4}+2}\
&=\frac{1}{2 + 2}\
&=\frac{1}{4}
\end{align*}
$$
# Answer:
The value of the limit is $ \frac{1}{4} $.

#### 4.
##### a.
# Explanation:
The average rate of change between two consecutive points is given by $ \frac{f(x_2)-f(x_1)}{x_2 - x_1} $, which is the slope of the secant line connecting the two points. To find which has the greatest average rate of change, we look at the steepest secant line. Visually, the secant line between the two points with the largest vertical change over the horizontal change will have the greatest slope. (Since the graph is not fully visible, but generally, we compare the slopes of secants between $ A - B $, $ B - C $, $ C - D $, $ D - E $).
##### b.
# Explanation:
The instantaneous rate of change at a point is the slope of the tangent line at that point. The average rate of change between $ A $ and $ B $ is the slope of the secant $ AB $. We need to check if there exists a point on the curve between $ A $ and $ B $ where the slope of the tangent (instantaneous rate) is greater than the slope of $ AB $ (average rate). By the Mean Value Theorem, if a function is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $ c\in(a,b) $ such that $ f^{\prime}(c)=\frac{f(b)-f(a)}{b - a} $. But if the function is not linear, there can be points where the tangent slope is greater or less. (Without the graph, we assume the function is differentiable and apply MVT concepts).
##### c.
# Explanation:
We need to sketch a tangent line at a point between $ D $ and $ E $ such that the slope of this tangent line is equal to the slope of the secant line between $ B $ and $ C $ (i.e., the average rate of change between $ B $ and $ C $). First, find the slope of $ BC $ (average rate of change $ \frac{f(C)-f(B)}{C - B} $), then find a point between $ D $ and $ E $ where the tangent to the curve has the same slope.

### PART B
#### 7.
##### a.
# Explanation:
The height function is $ s(t)=320 - 5t^2 $, $ 0\leq t\leq8 $. The average velocity over an interval $[a,b]$ is given by $ \frac{s(b)-s(a)}{b - a} $. For the first second, $ a = 0 $, $ b = 1 $.
First, find $ s(0) $ and $ s(1) $:
$ s(0)=320-5(0)^2 = 320 $
$ s(1)=320-5(1)^2=320 - 5=315 $
Then average velocity $ v_{avg}=\frac{s(1)-s(0)}{1 - 0}=\frac{315 - 320}{1}=- 5\space m/s $ (negative because height is decreasing)
##### b.
# Explanation:
For the interval from $ t = 1 $ to $ t = 3 $:
$ s(1)=315 $, $ s(3)=320-5(3)^2=320 - 45 = 275 $
Average velocity $ v_{avg}=\frac{s(3)-s(1)}{3 - 1}=\frac{275 - 315}{2}=\frac{- 40}{2}=-20\space m/s $
For the interval from $ t = 3 $ to $ t = 8 $:
$ s(3)=275 $, $ s(8)=320-5(8)^2=320 - 320 = 0 $
Average velocity $ v_{avg}=\frac{s(8)-s(3)}{8 - 3}=\frac{0 - 275}{5}=- 55\space m/s $
##### c.
# Explanation:
The velocity function is the derivative of the position function. $ s(t)=320-5t^2 $, so $ v(t)=s^{\prime}(t)=- 10t $.
At $ t = 2 $, $ v(2)=-10(2)=-20\space m/s $
The average velocity over $ 3\leq t\leq8 $ is what we calculated in part (b) as $ - 55\space m/s $ (wait, no, part (c) says "Calculate the average velocity for the interval $ 3\leq t\leq8 $". Wait, in part (b) we calculated for $ 1 - 3 $, $ 3 - 8 $. Wait, the question says: "c. Calculate the average velocity for the interval $ 3\leq t\leq8 $." We already did that in part (b) as $ \frac{s(8)-s(3)}{8 - 3}=\frac{0 - 275}{5}=- 55\space m/s $. And "Calculate the velocity at $ t = 2 $" is $ v(2)=- 20\space m/s $ as above.

# Answers:
#### 7a.
The average velocity during the first second is $ \boldsymbol{-5\space m/s} $
#### 7b.
- Average velocity during first second: $ - 5\space m/s $
- Average velocity during $ 1\leq t\leq3 $: $ \boldsymbol{-20\space m/s} $
- Average velocity during $ 3\leq t\leq8 $: $ \boldsymbol{-55\space m/s} $
#### 7c.
- Average velocity for $ 3\leq t\leq8 $: $ \boldsymbol{-55\space m/s} $
- Velocity at $ t = 2 $: $ \boldsymbol{-20\space m/s} $

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QUESTION IMAGE

Question

Explanation:

PART A

1.

Step 1: Recall when velocity is zero

Step 2: Solve for \( t \)

b.

3.

Answer:

2.

a.