Question

4.
The x and y-coordinates of a particle in motion, as
functions of time t, are given by: x= 712 - 4t +6,
y=3t9-312 - 12t-5(x and y are in mandt is in s.)
The x and y-components of the average velocity,
in the interval from t = 0 stot = 5 s are:​

Answer 1

Answer:

${\bold{\therefore Avg.\:velocity\:in\:x\:direction=31ms^{-1}}}$

${\bold{\therefore Avg.\:velocity\:in\:y\:direction=48ms^{-1}}}$

Explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about x and y coordinates of a particle in motion.

• Final and initial time is given.

• We have to find average velocity in both x and y direction.

$\underline \bold{Given : } \\ \bold{In \:horizontal \: direction }\\ \implies x = {7t}^{2} - 4t + 6 \\ \implies time(t) = 0 \: and \: 5 \\ \bold{In \:vertical \: direction} \\ \implies y = {3t}^{3} - 3 {t}^{2} - 12t - 5 \\ \implies time(t) = 0 \: and \: 5 \\ \\ \underline \bold{To \: Find : } \\ \implies Avg.\: velocity \: in \: x \: direction = ? \\ \implies Avg.\: velocity \: in \: y\: direction = ?$

• According to given question :

$\bold{In \: x \: direction : } \\ \implies x = {7t}^{2} - 4t + 6 \\ \bold{at \: t = 0 \: \: \: (Initial \: distance)} \\ \implies x_{i}= 7 \times {0}^{2} - 4 \times 0 + 6 \\ \bold{\implies x_{i} = 6 \: m} \\ \\ \bold{At \: t = 5 \: \: (final \: distance)} \\ \implies x_{f} = 7 \times {5}^{2} - 4 \times 5 + 6 \\ \implies x_{f} =7 \times 25 - 20 + 6 \\ \implies x_{f} =175 - 14 \\ \bold{\implies x_{f} =161 \: m} \\ \\ \implies Average \: velocity = \frac{ x_{f} - x_{i} }{ \triangle t} \\ \implies Average \: velocity = \frac{161 - 6}{5 - 0} \\ \implies Average \: velocity = \frac{155}{5} \\ \bold{\therefore Average \: velocity = 31 {m} {s}^{ - 1} }$

$\bold{In \: y \: direction : } \\ \implies y= {3t}^{3} - {3t}^{2} 12t - 5 \\ \bold{At \: t = 0 \: \: \: (Initial \: distance)} \\ \implies y_{i}= 3 \times {0}^{3} - {3 \times 0}^{2} \times 12 \times 0 + 6 \\ \bold{\implies y_{i} = - 5 \: m} \\ \\ \bold{At \: t = 5 \: \: (final \: distance)} \\ \implies y_{f} = 3 \times {5}^{3} - {3 \times 5}^{2} - 12 \times 5 - 5 \\ \implies y_{f} =3\times 125 - 75 - 60 - 5\\ \implies y_{f} =375 - 140 \\ \bold{\implies y_{f} =235 \: m} \\ \\ \implies Average \: velocity = \frac{ y_{f} - y_{i} }{ \triangle t} \\ \implies Average \: velocity = \frac{235 - ( - 5)}{5 - 0} \\ \implies Average \: velocity = \frac{240}{5} \\ \bold{\therefore Average \: velocity = 48 {m} {s}^{ - 1} }$

Answer 2

Solution:

Given:

➜ The x and y-coordinates of a particle in motion, as functions of time t, are given by: x = 7t² - 4t +6, y = 3t -3t² - 12t - 5(x and y are in mandt is in s.)

➜ The x and y-components of the average velocity,

in the interval from (t = 0) s to (t = 5) s are:

Find:

➜ Find the Average velocity of x-direction and y-direction.

Given:

Horizontal direction:

➜ x = 7t² - 4t + 6

➜ t = 0 and 5

Vertical direction:

➜ y = 3t³ - 3t² - 12t - 5

➜ t = 0 and 5

In x-direction:

➜ x = 7t² - 4t + 0

(t = 0) [ Initial distance ]

➜ xᵢ = 7 × 0² - 4 × 0 + 6

➜ xᵢ = 6 cm

(t = 5) [ final distance ]

➜ xƒ = 7 × 5² - 4 × 5 + 6

➜ xƒ = 7 × 25 - 20 + 6

➜ xƒ = 175 - 14

➜ xƒ = 161 m

➜ Avg.v = xƒ - xᵢ/Δ t

➜ Avg.v = 101 - 6/5 - 0

➜ Avg.v = 155/5

➜ Avg.v = 31 ms-¹

In y-direction:

➜ y =3t³ - 3t² 12t - 5

(t = 0) [ initial distance ]

➜ yᵢ = 3 × 0³ - 3 × 0² × 12 × 0 + 16

➜ yᵢ = - 5m

(t = 5) [ final distance ]

➜ yƒ = 3 × 5³ - 3 × 5² - 12 × 5 - 5

➜ yƒ = 3 × 125 - 75 - 60 - 5

➜ yƒ = 375 - 140

➜ yƒ 235 m

➜ Avg.v = yƒ - yᵢ/Δ t

➜ Avg.v = 235 - (-5)/5-0

➜ Avg.v = 248/5

➜ Avg.v = 40 ms-¹

Therefore, average velocity of x-direction = 31 ms-¹ and y-direction = 48 ms-1.

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