MCV4U Unit 2 Derivatives PDF

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MCV4U Unit 2: DerivativesAssessment of Learning1a. Given:𝑓(𝑥) = 13𝑥 4 − 7𝑥 3 + 5𝑥 2 + 11𝑥 + 75𝑓′(𝑥) = 4(13𝑥 3 ) − 3(7𝑥 2 ) + 2(5𝑥) + 1(11) + 0𝑓′(𝑥) = 52𝑥3 − 21𝑥 2 + 10𝑥 + 111b. Given:𝑓(𝑥) = (𝑥 3 + 2𝑥 2 + 4)(𝑥 4 − 3)𝑓(𝑥) = 𝑥3 + 2𝑥 2 + 4𝑓′(𝑥) = 3𝑥 2 + 4𝑥𝑔(𝑥) = 𝑥 4 − 3𝑔′(𝑥) = 4𝑥 3𝐹′(𝑥) = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(...

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MCV4U Unit 2: Derivatives Assessment of Learning 1a. Given: 4

𝑓(𝑥) = 13𝑥

3

2

− 7𝑥

+ 5𝑥

3

+ 11𝑥 + 75

2

𝑓′(𝑥) = 4(13𝑥 ) − 3(7𝑥 ) + 2(5𝑥) + 1(11) + 0 3

𝑓′(𝑥) = 52𝑥

2

− 21𝑥

+ 10𝑥 + 11

1b. Given: 3

2

𝑓(𝑥) = (𝑥

4

+ 2𝑥

3

+ 4)(𝑥

2

𝑓(𝑥) = 𝑥

+ 2𝑥 2

𝑓′(𝑥) = 3𝑥 4

𝑔(𝑥) = 𝑥

− 3)

+ 4

+ 4𝑥

− 3 3

𝑔′(𝑥) = 4𝑥 𝐹′(𝑥) = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥) 2

𝐹′(𝑥) = (3𝑥 6

𝐹′(𝑥) = 3𝑥

4

+ 4𝑥)(𝑥 2

− 9𝑥

3

− 3) + (𝑥 5

+ 4𝑥

2

+ 2𝑥 6

− 12𝑥 + 4𝑥

3

+ 4)(4𝑥 ) 5

3

+ 8𝑥 + 16𝑥

6

𝐹′(𝑥) = 7𝑥

5

3

+ 12𝑥

+ 16𝑥

1c. Given: 2

𝑓(𝑥) =

3𝑥 − 6𝑥 + 7 4𝑥−1 2

𝑓(𝑥) = 3𝑥 − 6𝑥 + 7 𝑓'(𝑥) = 6𝑥 − 6 𝑔(𝑥) = 4𝑥 − 1 𝑔'(𝑥) = 4 𝐹'(𝑥) =

𝑓'(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔'(𝑥) [𝑔(𝑥)]

2 2

𝐹'(𝑥) =

(6𝑥−6)(4𝑥−1) − (3𝑥 −6𝑥+7)(4)

𝐹'(𝑥) =

24𝑥 −6𝑥−24𝑥+6−12𝑥 +24𝑥−28

𝐹'(𝑥) =

12𝑥 −6𝑥−22

2

(4𝑥−1) 2

2

2

(4𝑥−1) 2

2

(4𝑥−1)

1d. Given: 3

𝑓(𝑥) = (4𝑥 − 7𝑥)

7

2

− 9𝑥

− 12𝑥

3

6

2

𝑓'(𝑥) = 7(4𝑥 − 7𝑥) (12𝑥 − 7)

1e. Given: 2

2

12 = 𝑦 + 𝑥 2

2

𝑥 +𝑦 = 0 2𝑥 + 2𝑦(𝑦') = 0 2𝑦(𝑦') = − 2𝑥 𝑦' = 𝑦' =

−2𝑥 2𝑦 −𝑥 𝑦

2

2

𝑦 = 12 − 𝑥 𝑦= 𝑦' =

2

12 − 𝑥 −𝑥 2

12−𝑥

1f. Given: 𝑓(𝑥) = 156

𝑓'(𝑥) = 0

1g. Given: 2

2

𝑓(𝑥) = (1 − 3𝑥) (𝑥 − 2)

3

2

𝑓(𝑥) = (1 − 3𝑥 ) 𝑓'(𝑥) = [2(1 − 3𝑥)][− 3] = (− 3)(2 − 6𝑥) = 2

𝑔(𝑥) = (𝑥 − 2)

− 6 + 18𝑥

3

2

2

2

2

2

3

𝑔'(𝑥) = [3(𝑥 − 2) ][2𝑥] = 2𝑥(3(𝑥 − 2) ) 5

3

𝑔'(𝑥) = 6𝑥(𝑥 − 2) = 6𝑥 − 24𝑥 + 24𝑥 𝐹'(𝑥) = 𝑓'(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔'(𝑥) 2

3

2

5

3

𝐹'(𝑥) = (− 6 + 18𝑥)(𝑥 − 2) + (1 − 3𝑥) (6𝑥 − 24𝑥 + 24𝑥) 1h. Given:

2

𝑓(𝑥) = (

2𝑥 +1 2

)

3

3𝑥 +1

𝑓(𝑥) =

4

2

6

3

4𝑥 +4𝑥 +1 9𝑥 +6𝑥 +1 4 2

𝑓(𝑥) = 4𝑥 + 4𝑥 + 1 3

3

𝑓'(𝑥) = 4(4𝑥 ) + 2(4𝑥) = 16𝑥 + 8𝑥 6

3

𝑔(𝑥) = 9𝑥 + 6𝑥 + 1 5

2

5

2

𝑔'(𝑥) = 6(9𝑥 ) + 3(6𝑥 ) = 54𝑥 + 18𝑥 𝐹'(𝑥) =

𝑓'(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔'(𝑥)

𝐹'(𝑥) =

(16𝑥 +8𝑥)(9𝑥 +6𝑥 +1) − (4𝑥 +4𝑥 +1)(54𝑥 +18𝑥 )

𝐹'(𝑥) =

144𝑥 +72𝑥 +96𝑥 +48𝑥 +16𝑥 +8𝑥−216𝑥 −216𝑥 −72𝑥 −54𝑥 −72 −18𝑥

𝐹'(𝑥) =

−72𝑥 −144𝑥 +24𝑥 −54𝑥 −24𝑥 +16𝑥 −18𝑥 +8𝑥

[𝑔(𝑥)]

2

3

6

3

4

6

2

3

5

2

2

(9𝑥 +6𝑥 +1) 9

7

6

4

3

9

6

3

7

2

(9𝑥 +6𝑥 +1) 9

7

6

5

6

4

3

3

2

(9𝑥 +6𝑥 +1)

2. Given: 2

𝑓(𝑥) = (3𝑥 − 7𝑥 + 4) 2

8

7

𝑓'(𝑥) = [8(3𝑥 − 7𝑥 + 4) ][2(3𝑥) − 7] 2

7

𝑓'(𝑥) = (24𝑥 − 56𝑥 + 32) (6𝑥 − 7)

2

6

5

4

2

2

6

𝑓''(𝑥) = [7(24𝑥 − 56𝑥 + 32) (2(24𝑥)) − 56)][6] 2

6

2

6

𝑓''(𝑥) = (168𝑥 − 392𝑥 + 224) (48𝑥 − 56)(6) 𝑓''(𝑥) = (168𝑥 − 392𝑥 + 224) (288𝑥 − 336) 3. Given: 10

𝑓(𝑥) = 12𝑥

7

5

2

− 3𝑥 + 4𝑥 + 5𝑥 + 6 9

6

4

𝑓'(𝑥) = 10(12𝑥 ) − 7(3𝑥 ) + 5(4𝑥 ) + 2(5𝑥) + 0 9

6

4

𝑓'(𝑥) = 120𝑥 − 21𝑥 + 20𝑥 + 10𝑥 8

5

3

𝑓''(𝑥) = 9(120𝑥 ) − 6(21𝑥 ) + 4(20𝑥 ) + 10 8

5

3

𝑓''(𝑥) = 1080𝑥 − 126𝑥 + 80𝑥 + 10 7

4

2

𝑓'''(𝑥) = 8(1080𝑥 ) − 5(126𝑥 ) + 3(80𝑥 ) + 0 7

4

2

𝑓'''(𝑥) = 8640𝑥 − 630𝑥 + 240𝑥

4. Given: 2

𝑦 = 𝑥 − 2𝑥 + 3 Point (2,3) I first found the derivative and slope of the function:

𝑓'(𝑥) = 2𝑥 − 2 𝑚 = 𝑓'(2) 𝑓'(2) = 2(2) − 2 = 2 𝑚= 2 I then found the equation of the tangent line to the curve: 𝑦 = 𝑚(𝑥 − 𝑥1) + 𝑦1 𝑦 = 2(𝑥 − 2) + 3 𝑦 = 2𝑥 − 4 + 3 𝑦 = 2𝑥 − 1 Therefore, the equation of the tangent line to the curve 2

𝑦 = 𝑥 − 2𝑥 + 3at the point (2,3) is𝑦 = 2𝑥 − 1.

5. Given: 2

𝑦 = − 𝑥 + 3𝑥 + 4 Slope = 5

I first found the derivative of the function, then substituted the slope into the derivative to find the x value, and lastly substituted in the x value into the function to find the y value: 𝑓'(𝑥) = 2(− 𝑥) + 3 + 0 𝑓'(𝑥) = − 2𝑥 + 3 𝑚= 5 5 = − 2𝑥 + 3 − 2𝑥 = 2 𝑥= − 1 2

𝑦 = − (− 1) + 3(− 1) + 4 𝑦= − 1 − 3 + 4 𝑦= 0 Therefore, the slope of the tangent line is 5 on the 2

parabola 𝑦 =− 𝑥 + 3𝑥 + 4 at the point (-1,0).

6. One example of a rate that is decreasing is a population decreasing by 40 per week.

One example of a rate that is increasing is a car driving at a speed of 40 miles per hour. The rate that is increasing here is the distance the car has travelled. 7. When a vehicle is said to be travelling at -35km/h, it infers that the vehicle is travelling backwards at 35km/h. 8. 𝑑𝑉 represents 𝑑𝑡

a change in volume with respect to a

change in time.

9a. Given: 2

𝑠(𝑡) = 90 − 4. 9𝑡

𝑡> 0 I first found the velocity formula by finding the derivative of the given function, and then substituted in 1 and 4 to find the velocity of the hammer at that time: 𝑠'(𝑡) = 𝑣(𝑡) = 2(− 4. 9𝑡) = − 9. 8𝑡 𝑣(𝑡) = − 9. 8𝑡 𝑣(1) = − 9. 8(1) = − 9. 8𝑚 𝑣(4) = − 9. 8(4) = − 39. 2 Therefore, at 1 second, the velocity of the hammer is 9.8m/s downwards, and at 4 seconds, the velocity of the hammer is 39.2m/s downwards.

9b. Given:

2

𝑠(𝑡) = 90 − 4. 9𝑡 𝑡> 0

The hammer will have a height of 0 when it is on the ground, so I set 𝑠 = 0 and solved for 𝑡 to find the approximate time that the hammer will hit the ground: 2

0 = 90 − 4. 9𝑡 2

4. 9𝑡 = 90 𝑡=

90 4.9

𝑡 ≈ 4. 3 Therefore, the hammer will hit the ground at approximately 4.3 seconds.

9c. Given: 2

𝑠(𝑡) = 90 − 4. 9𝑡 𝑡> 0

I used the previously found velocity formula and the time the hammer hits the ground to find the impact velocity of the hammer: 𝑣(𝑡) = − 9. 8𝑡 𝑣(4. 3) = − 9. 8(4. 3) 𝑣(4. 3) = − 42. 14𝑚/𝑠 Therefore, the impact velocity of the hammer is -42.14/ms.

10a. Given: 3

2

𝑠(𝑡) = 2𝑡 − 15𝑡 + 33𝑡 + 17 𝑡> 0 I first found the velocity formula by finding the derivative of the given function, and set it equal to 0 because the particle will be at rest at 0: 2

𝑠'(𝑡) = 𝑣(𝑡) = 3(2𝑡 ) − 2(15𝑡) + 33 2

𝑣(𝑡) = 6𝑡 − 30𝑡 + 33 2

2

0 = 6𝑡 − 30𝑡 + 33 3(2𝑡 − 10𝑡 + 11) = 0 In order to fully solve this, I used the quadratic formula: 𝑎 = 2, 𝑏 = 10, 𝑐 = 11 2

𝑥=

10± (−10) −4(2)(11) 2(2)

𝑥=

10± (100−88 4

𝑥=

10± 12 4

𝑥 ≈ 3. 4 𝑜𝑟 𝑥 ≈ 1. 6 Therefore, the particle is at rest at approximately 3.4 seconds and 1.6 seconds.

10b. I know the particle is at rest at approximately 3.4 and 1.6 seconds, this means that the particle must be moving in one direction before or after these times. Thus, I created a table to find the when the velocity would be positive: 2

Interval

Point

𝑣(𝑡) = 6𝑡 − 30𝑡 − 33

0 < 𝑡 < 1. 6

1

2 𝑣(1) = 6(1) − 30(1) + Positive = 6 − 30 + 33 = 9

1. 6 < 𝑡 < 3. 4 3

𝑡 > 3. 4

5

Direction

2 𝑣(3) = 6(3) − 30(3) + Negative = 54 − 90 + 33 = − 3 2 𝑣(5) = 6(5) − 30(5) + Positive = 150 − 150 + 33 = 33

Therefore, the velocity is positive between 1 and 1.6 seconds, and again after 3.4 seconds.

10c. First, I found what displacement is at 0 seconds: 3

2

𝑠(𝑡) = 2𝑡 − 15𝑡 + 33𝑡 + 17 3

2

𝑠(0) = 2(0) − 15(0) + 33(0) + 17 𝑠(0) = 17 Therefore, the particle is 17 meters from the reference point. This means that for the next 1.6 seconds, the particle is moving forwards. I then found where the particle stops at 1.6 seconds using the displacement formula: 3

2

𝑠(𝑡) = 2𝑡 − 15𝑡 + 33𝑡 + 17 3

2

𝑠(1. 6) = 2(1. 6) − 15(1. 6) + 33(1. 6) + 17 𝑠(1. 6) = 39. 6 Therefore, at 1.6 seconds, the displacement is 39.6m. This means that the particle has moved forwards 22.6m because displacement is always to the reference point. However, in the interval of 1.6 to 3.4 seconds the particle is moving backwards. I then found the displacement at 3.4 seconds:

3

2

𝑠(𝑡) = 2𝑡 − 15𝑡 + 33𝑡 + 17 3

2

𝑠(3. 4) = 2(3. 4) − 15(3. 4) + 33(3. 4) + 17 𝑠(3. 4) = 34. 4 Therefore, at 3.4 seconds, the particle is being displaced by 34.4m, which means it is 34.4m from the reference point. This also means that the particle has moved backwards 5.2m because the particle at 1.6s was displaced at 39.6m and at 3.4 seconds it was displaced at 34.4m. Thus, in the interval of 3.4 seconds to 10 seconds, the particle is moving forward again. I then found the displacement at 10 seconds: 3

2

𝑠(𝑡) = 2𝑡 − 15𝑡 + 33𝑡 + 17 3

2

𝑠(10) = 2(10) − 15(10) + 33(10) + 17 𝑠(10) = 847 Therefore, at 10 seconds, the particle is being displaced 847m, which means that it is 847m from the reference point. This also means that the particle has moved forward 812.6m because the particle was displaced 34.4m at 3.4s and 847m at 10s. Diagram:

10d. During the first 1.6 seconds the particle moves forward 22.6m. In the following 1.8 seconds, the particle moves backwards 5.2m. In the following 6.6 seconds, the particle moves forwards 812.6m. 22. 6 + 5. 2 + 812. 6 = 839. 8𝑚 Therefore, the particle travelled a total distance of 839.8m in the first 10 seconds. 11. Given: 𝑟 = 4𝑐𝑚 𝑑𝑆𝐴 𝑑𝑡

=

2

− 0. 5𝑐𝑚 /𝑚𝑖𝑛 2

𝑆𝐴 = 4π𝑟

I first found the derivative, and then solved for 𝑑𝑆𝐴 𝑑𝑡

2

= (4𝑟 )(2𝑟)

𝑑𝑟 𝑑𝑡

:

𝑑𝑆𝐴 𝑑𝑡

2

= (4𝑟 )(2𝑟)

𝑑𝑟 𝑑𝑡

− 0. 5 = (4π)(2(4)) − 0. 5 = (4π)(8) − 0. 5 = 32π 𝑑𝑟 𝑑𝑡 𝑑𝑟 𝑑𝑡

𝑑𝑟 𝑑𝑡

𝑑𝑟 𝑑𝑡

𝑑𝑟 𝑑𝑡

0.5 32π

=

−

=

− 0. 00497𝑐𝑚/𝑚𝑖𝑛

Therefore, the radius of the snowball decreases at a rate of -0.00497cm/min.

12. Given: 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡

=

− 60𝑘𝑚/ℎ

=

− 70𝑘𝑚/ℎ

𝑥 = 0. 4𝑘𝑚 𝑦 = 0. 3𝑘𝑚 I first found the z value using pythagorean theorem: 2

2

2

𝑧 =𝑥 +𝑦

2

2

𝑧 = (0. 4) + (0. 3)

2

2

𝑧 = 0. 16 + 0. 09 𝑧 = 0. 5 I then found the derivative, substituted in my values, and solved to find the rate of decrease: 𝑑𝑧 𝑑𝑡

2𝑧

= 2𝑥

2(0. 5) 𝑑𝑧 𝑑𝑡 𝑑𝑧 𝑑𝑡

𝑑𝑧 𝑑𝑡

𝑑𝑥 𝑑𝑡

+ 2𝑦

𝑑𝑦 𝑑𝑡

= 2(0. 4)(− 60) + 2(0. 3)(− 70)

=

− 48 − 42

=

− 90𝑘𝑚/ℎ

Therefore, the distance between the cars is decreasing at a rate of 90km/h. 13. 3

𝑑𝑉 𝑑𝑡

= 1. 2𝑚 /𝑚𝑖𝑛

ℎ = 5. 2𝑚 𝑑 = 5𝑚 𝑟 = 2. 5𝑚 I first created a ratio for radius and height, so I could rearrange it and have a specific relationship for the radius: 𝑟: ℎ = 2. 5: 5. 2 𝑟 ℎ

= 𝑟

2.5 5.2 2.5

ℎ( ℎ ) = ℎ( 5.2 )

2.5ℎ 5.2

𝑟=

I then substituted this radius expression into the velocity formula: 𝑉 =

π(

2.5ℎ 2 )ℎ 5.2

3 2

𝑉 =

π(

6.25π 27.04

)ℎ

3 6.25πℎ 27.04

𝑉 =

3

3 3

𝑉 =

6.25πℎ 81.12

3

Next, I found the height when there is 8π𝑚 amount of water: 3

6.25πℎ 81.12 3

8π =

6. 25πℎ = (8π)(81. 12) 3

(8π)(81.12) 6.25π

ℎ =

ℎ = 4. 7 I then found the derivative: 3

𝑉 = 𝑑𝑉 𝑑𝑡 𝑑𝑉 𝑑𝑡

6.25π 3 ℎ 81.12 2 𝑑ℎ 6.25π (3ℎ ) 𝑑𝑡 81.12

6.25πℎ 81.12

=

=

2

=

18.75πℎ 81.12

𝑑ℎ 𝑑𝑡

I then substituted in the h value to find height change:

𝑑𝑉 𝑑𝑡

2

=

𝑑ℎ 𝑑𝑡

18.75πℎ 81.12

2

1. 2 =

18.75π(4.7) 81.12 1.2

𝑑ℎ 𝑑𝑡

=

𝑑ℎ 𝑑𝑡

= 0. 07

18.75(4.7) 91.12

𝑑ℎ 𝑑𝑡

2

Therefore, the water level is rising at a rate of 0.07m/min 3

when there is 8π𝑚 amount of water. 14. Given: 3

2

3

2

𝑓(𝑥) = 𝑎𝑥 + 𝑏𝑥 − 5𝑥 + 9 I first substituted in -1 to the function, because that is the value we are trying to find a and b of: 𝑓(𝑥) = 𝑎𝑥 + 𝑏𝑥 − 5𝑥 + 9 3

2

𝑓(− 1) = 𝑎(− 1) + 𝑏(− 1) − 5(− 1) + 9 12 = − 𝑎 + 𝑏 + 14 𝑏 = − 2 + 𝑎 I then took the derivative of the function and substituted in -1 again to get a formula for a: 3

2

𝑓(𝑥) = 𝑎𝑥 + 𝑏𝑥 − 5𝑥 + 9 2

𝑓'(𝑥) = 3(𝑎𝑥 ) + 2(𝑏𝑥) − 5 + 0 2

𝑓'(𝑥) = 3𝑎𝑥 + 2𝑏𝑥 − 5 2

𝑓'(− 1) = 3𝑎𝑥 + 2𝑏𝑥 − 5

2

3 = 3𝑎(− 1) + 2𝑏(− 1) − 5 3 = 3𝑎 − 2𝑏 − 5 8 = 3𝑎 − 2𝑏 I then substituted the b value to find the value of a: 8 = 3𝑎 − 2𝑏 8 = 3𝑎 − 2(− 2 + 𝑎) 8 = 3𝑎 + 4 − 2𝑎 𝑎 = 4 I then substituted the a value to find the value of b: 𝑏 = − 2 + 𝑎 𝑏 = − 2 + 4 𝑏 = 2 Therefore, the values of a and b so that f(-1)=12 and f’(-1)=3 are a=4 and b=2. 15. Given: 2

𝑦=𝑥 + 1 Point (-1,-3) I first found the derivative and substituted in the values of the given point: 𝑦' = 2𝑥 𝑚 = 2𝑥 𝑚=

𝑦2−𝑦1 𝑥2−𝑥1

2

2𝑥 = 2𝑥 =

(𝑥 +1) − (−3) 𝑥 − (−1) 2 𝑥 +4 𝑥+1

2

=

𝑥 +1+3 𝑥+1

2

2𝑥(𝑥 + 1) = 𝑥 + 4 2

2

2𝑥 + 2𝑥 = 𝑥 + 4 2

𝑥 + 2𝑥 − 4 = 0 I fully solved this using the quadratic formula: 𝑎 = 1, 𝑏 = 2, 𝑐 = − 4 2

𝑥=

−2± (2) −4(1)(−4) 2(1)

𝑥=

−2± 20 2

𝑥 ≈ 1. 236 𝑥 ≈ − 3. 236 I then used these x-values to make the equation of the two lines: First Equation: Point: 2

𝑦=𝑥 + 1 2

𝑦 = (1. 236) + 1 = 2. 528 𝑥1, 𝑦1(1. 236, 2. 528) Slope: 𝑚 = 2𝑥 𝑚1 = 2(1. 236)

𝑚1 = 2. 472 Equation of a tangent line: 𝑦 = 𝑚(𝑥 = 𝑥1) + 𝑦1 𝑦 = 2. 472(𝑥 − 1. 236) + 2. 528 𝑦1 = 2. 472𝑥 − 0. 527 Second Equation: Point: 2

𝑦=𝑥 + 1 2

𝑦 = (− 3. 236) + 1 = 11. 472 𝑥2, 𝑦2(− 3. 236, 11. 472) Slope: 𝑚 = 2𝑥 𝑚2 = 2(− 3. 236) 𝑚2 =

− 6. 472

Equation of a tangent line: 𝑦 = 𝑚(𝑥 = 𝑥1) + 𝑦1 𝑦 = − 6. 472(𝑥 + 3. 236) + 11. 472 𝑦2 = − 6. 472𝑥 − 9. 471 Therefore, the equations of two lines that pass through the 2

point (-1,-3) and are tangent to the graph of 𝑦 = 𝑥 + 1are 𝑦1 = 2. 472𝑥 − 0. 527and 𝑦2 = − 6. 472𝑥 − 9. 471

16. Given: 2

𝑠1 = 4𝑡 − 𝑡 2

3

𝑠2 = 5𝑡 − 𝑡

I first took the derivative of the first displacement formula to find the acceleration: 2

𝑠1 = 4𝑡 − 𝑡

𝑣1 = 4 − 2𝑡 𝑎1 =

− 2

I did the same with second displacement formula: 2

3

𝑠2 = 5𝑡 − 𝑡

2

2

𝑣2 = 2(5𝑡) − 3𝑡 = 10𝑡 − 3𝑡 𝑎2 = 10 − 6𝑡

I then set both of the acceleration values equal to each other to find the value of time: − 2 = 10 − 6𝑡 6𝑡 = 12 𝑡= 2

I then substituted in the value of t into the two velocity formulas to find my final answer: 𝑣1 = 4 − 2𝑡 𝑣1 = 4 − 2(2) 𝑣1 = 0 2

𝑣2 = 10𝑡 − 3𝑡

𝑣2 = 10(2) − 3(2)

2

𝑣2 = 20 − 12 𝑣2 = 8 Therefore, 𝑣1 = 0 𝑣2 = 8 when the accelerations of the two particles are equal....