Question

The x and y-coordinates of a particle in motion,as functions of time t,are given by: x=7t^2-4t+6, y=3t^3-3t^2-12t-5(x and y are in m and t is in s.) Then what are the x and y-components of the average velocity,in the interval from t= 0 s to t= 5 s??

Answer 1

Answer:

x component of average velocity = 31 m/s.

y component of average velocity = 48 m/s.

Explanation:

Given:

The x and y coordinates of the particle,

• $x=7t^2-4t+6.$

• $y=3t^3-3t^2-12t-5$

The average velocity of the particle in the time interval from initial time $t_i$ to final time $t_f$ is defined as

$v_{av} = \dfrac{r_f-r_i}{t_f-t_i}.$

where,

$r_i,\ r_f$ are the initial and final positions of the particle at time $t_i$ and $t_f$ respectively.

For the x component of average velocity,

$v_{avx}=\dfrac{x_f-x_i}{t_f-t_i}$

where,

$t_i=0\ s\\t_f=5\ s.\\\therefore x_i = (x)_{t=0}=(7t^2-4t+6)_{t=0} = (7\cdot 0^2-4\cdot 0+6)=6\ m.\\\text{and}\ x_f=(x)_{t=5}=(7t^2-4t+6)_{t=5} = (7\cdot 5^2-4\cdot 5+6)=161\ m.\\\\v_{avx} = \dfrac{161-6}{5-0}=\dfrac{155}{5} = 31\ m/s.$

For the y component of average velocity,

$v_{avy}=\dfrac{y_f-y_i}{t_f-t_i}$

where,

$t_i=0\ s\\t_f=5\ s.\\\therefore y_i = (y)_{t=0}=(3t^3-3t^2-12t-5)_{t=0} = (3\cdot 0^3-3\cdot 0^2-12\cdot 0-5)=-5\ m.\\\text{and}\ y_f=(y)_{t=5}=(3t^3-3t^2-12t-5)_{t=0} = (3\cdot 5^3-3\cdot 5^2-12\cdot 5-5)=235\ m\\\\v_{avy} = \dfrac{235-(-5)}{5-0}=\dfrac{240}{5} = 48\ m/s.$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

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

$\green{\therefore {\text{Avg.\:velocity\:in\:y\:direction=48m}s^{-1}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

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

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.

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

• According to given question :

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

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