day 3: rates of change - worksheet
1 the velocity of an object is given by ( v(t) = t(t - 4)^2 ). at what times, in seconds, is the object at rest?
2 a construction worker drops a bolt while working on a high - rise building, 320 m above the ground. after ( t ) seconds, the bolt has fallen a distance of ( s ) metres, where ( s(t)=320 - 5t^2 ), ( 0leq tleq8 ).
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 ).
3 the function ( s(t)=8t(t + 2) ) describes the distance ( s ), in kilometres, that a car has travelled after a time ( t ), in hours, for ( 0leq tleq5 ).
a. calculate the average velocity of the car during the following intervals:
i. from ( t = 3 ) to ( t = 4 )
ii. from ( t = 3 ) to ( t = 3.1 )
iii. from ( t = 3 ) to ( t = 3.01 )
b. use your results for part a to approximate the instantaneous velocity of the car at ( t = 3 ).
c. calculate the velocity at ( t = 3 ).

Question

day 3: rates of change - worksheet
1 the velocity of an object is given by ( v(t) = t(t - 4)^2 ). at what times, in seconds, is the object at rest?
2 a construction worker drops a bolt while working on a high - rise building, 320 m above the ground. after ( t ) seconds, the bolt has fallen a distance of ( s ) metres, where ( s(t)=320 - 5t^2 ), ( 0leq tleq8 ).
   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 ).
3 the function ( s(t)=8t(t + 2) ) describes the distance ( s ), in kilometres, that a car has travelled after a time ( t ), in hours, for ( 0leq tleq5 ).
   a. calculate the average velocity of the car during the following intervals:
      i. from ( t = 3 ) to ( t = 4 )
      ii. from ( t = 3 ) to ( t = 3.1 )
      iii. from ( t = 3 ) to ( t = 3.01 )
   b. use your results for part a to approximate the instantaneous velocity of the car at ( t = 3 ).
   c. calculate the velocity at ( t = 3 ).

Accepted Answer

### Problem 1: Velocity of an Object
The velocity function is $ v(t) = t(t - 4)^2 $. We need to find when the object is at rest, i.e., when $ v(t)=0 $.

#### Step 1: Set the velocity function to zero
We have the equation $ t(t - 4)^2=0 $.

#### Step 2: Solve for $ t $
Using the zero - product property, if $ ab = 0 $, then either $ a = 0 $ or $ b=0 $.
- For $ t=0 $, the equation is satisfied since when $ t = 0 $, $ 0\times(0 - 4)^2=0\times16 = 0 $.
- For $ (t - 4)^2=0 $, taking the square root of both sides, we get $ t-4=0 $, so $ t = 4 $.

### Problem 2: Bolt Dropped from a Building
The distance function is $ s(t)=320 - 5t^2 $, $ 0\leq t\leq8 $

#### Part (a): Average velocity during the first, third, and eighth seconds
The average velocity over the interval $[a,b]$ is given by $ \text{Average velocity}=\frac{s(b)-s(a)}{b - a} $

- **First second ($ t = 0 $ to $ t = 1 $)**
    - Step 1: Find $ s(0) $ and $ s(1) $
      - $ s(0)=320-5\times0^2=320 $
      - $ s(1)=320 - 5\times1^2=320 - 5=315 $
    - Step 2: Calculate average velocity
      - $ \text{Average velocity}=\frac{s(1)-s(0)}{1 - 0}=\frac{315 - 320}{1}=- 5\space m/s $
- **Third second ($ t = 2 $ to $ t = 3 $)**
    - Step 1: Find $ s(2) $ and $ s(3) $
      - $ s(2)=320-5\times2^2=320 - 20 = 300 $
      - $ s(3)=320-5\times3^2=320 - 45 = 275 $
    - Step 2: Calculate average velocity
      - $ \text{Average velocity}=\frac{s(3)-s(2)}{3 - 2}=\frac{275 - 300}{1}=-25\space m/s $
- **Eighth second ($ t = 7 $ to $ t = 8 $)**
    - Step 1: Find $ s(7) $ and $ s(8) $
      - $ s(7)=320-5\times7^2=320 - 245 = 75 $
      - $ s(8)=320-5\times8^2=320 - 320 = 0 $
    - Step 2: Calculate average velocity
      - $ \text{Average velocity}=\frac{s(8)-s(7)}{8 - 7}=\frac{0 - 75}{1}=-75\space m/s $

#### Part (b): Average velocity for the interval $ 3\leq t\leq8 $
- Step 1: Find $ s(3) $ and $ s(8) $
    - $ s(3)=320-5\times3^2=320 - 45 = 275 $
    - $ s(8)=0 $
- Step 2: Calculate average velocity
    - $ \text{Average velocity}=\frac{s(8)-s(3)}{8 - 3}=\frac{0 - 275}{5}=\frac{- 275}{5}=-55\space m/s $

#### Part (c): Velocity at $ t = 2 $
The velocity function is the derivative of the distance function. If $ s(t)=320 - 5t^2 $, then $ v(t)=s^\prime(t) $
- Step 1: Differentiate $ s(t) $
    - Using the power rule $ \frac{d}{dt}(t^n)=nt^{n - 1} $, $ s^\prime(t)=\frac{d}{dt}(320)-5\frac{d}{dt}(t^2) $
    - $ \frac{d}{dt}(320) = 0 $ and $ \frac{d}{dt}(t^2)=2t $, so $ v(t)=- 10t $
- Step 2: Evaluate at $ t = 2 $
    - $ v(2)=-10\times2=-20\space m/s $

### Problem 3: Car's Distance Function $ s(t)=8t(t + 2) $

#### Part (a): Average velocity over intervals
The average velocity over $[a,b]$ is $ \frac{s(b)-s(a)}{b - a} $, and $ s(t)=8t^2+16t $

- **i. From $ t = 3 $ to $ t = 4 $**
    - Step 1: Find $ s(3) $ and $ s(4) $
      - $ s(3)=8\times3^2+16\times3=8\times9 + 48=72 + 48 = 120 $
      - $ s(4)=8\times4^2+16\times4=8\times16+64 = 128+64 = 192 $
    - Step 2: Calculate average velocity
      - $ \text{Average velocity}=\frac{s(4)-s(3)}{4 - 3}=\frac{192 - 120}{1}=72\space km/h $
- **ii. From $ t = 3 $ to $ t = 3.1 $**
    - Step 1: Find $ s(3) $ and $ s(3.1) $
      - $ s(3)=120 $ (from above)
      - $ s(3.1)=8\times(3.1)^2+16\times3.1=8\times9.61+49.6=76.88 + 49.6 = 126.48 $
    - Step 2: Calculate average velocity
      - $ \text{Average velocity}=\frac{s(3.1)-s(3)}{3.1 - 3}=\frac{126.48 - 120}{0.1}=\frac{6.48}{0.1}=64.8\space km/h $
- **iii. From $ t = 3 $ to $ t = 3.01 $**
    - Step 1: Find $ s(3) $ and $ s(3.01) $
      - $ s(3)=120 $
      - $ s(3.01)=8\times(3.01)^2+16\times3.01=8\times9.0601+48.16=72.4808+48.16 = 120.6408 $
    - Step 2: Calculate average velocity
      - $ \text{Average velocity}=\frac{s(3.01)-s(3)}{3.01 - 3}=\frac{120.6408 - 120}{0.01}=\frac{0.6408}{0.01}=64.08\space km/h $

#### Part (b): Approximate instantaneous velocity at $ t = 3 $
As the time interval $[3,b]$ gets smaller (i.e., $ b
ightarrow3 $), the average velocity over $[3,b]$ approaches the instantaneous velocity at $ t = 3 $. From part (a) ii and iii, as the interval length decreases (from $ 0.1 $ to $ 0.01 $), the average velocity approaches $ 64\space km/h $ (the exact value can be found by taking the derivative)

#### Part (c): Calculate velocity at $ t = 3 $
First, find the derivative of $ s(t) $. Since $ s(t)=8t^2 + 16t $
- Step 1: Differentiate $ s(t) $
    - Using the power rule, $ s^\prime(t)=16t + 16 $
- Step 2: Evaluate at $ t = 3 $
    - $ s^\prime(3)=16\times3+16=48 + 16=64\space km/h $

### Answers

#### Problem 1
The object is at rest at $ t = 0 $ seconds and $ t = 4 $ seconds.

#### Problem 2
- (a)
    - First second: $-5\space m/s$
    - Third second: $-25\space m/s$
    - Eighth second: $-75\space m/s$
- (b) $-55\space m/s$
- (c) $-20\space m/s$

#### Problem 3
- (a)
    - i. $ 72\space km/h$
    - ii. $ 64.8\space km/h$
    - iii. $ 64.08\space km/h$
- (b) Approximately $ 64\space km/h$
- (c) $ 64\space km/h$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1: Velocity of an Object

Step 1: Set the velocity function to zero

Step 2: Solve for \( t \)

Problem 2: Bolt Dropped from a Building

Part (a): Average velocity during the first, third, and eighth seconds

Part (b): Average velocity for the interval \( 3\leq t\leq8 \)

Part (c): Velocity at \( t = 2 \)

Problem 3: Car's Distance Function \( s(t)=8t(t + 2) \)

Part (a): Average velocity over intervals

Answer:

Problem 1

Problem 2

Problem 3