Question

Answer correctly or i'll report u

Answer 1

$\huge\bold{\underline{\underline{Given:-}}}$

• Displacement of a body along a straight line is described by the equation is x = 2t³ + 4t - 10
• (Here i have mentioned the term as displacement because the body is moving along strainght line which is nothing but the shortest path of the motion)

$\huge\bold{\underline{\underline{To\:Find:-}}}$

• Velocity and acceleration of the body at t = 4 sec

$\huge\bold{\underline{\underline{Solution:-}}}$

We know that ,

$\huge\bold \star \: \purple { \bold{ \boxed{ v \: = \: \frac{dx}{dt} }}}$

Where ,

• V is velocity
• x is displacement
• t is time

$\implies \: v \: = \: \frac{d}{dt} (2t {}^{3} + 4t - 10) \\ \\ \implies \: v = \frac{d}{dt} (2t {}^{3} ) + \frac{d}{dt} (4t) - \frac{d}{dt} (10)$

$\bold {\boxed{ \frac{d}{dx} (a {}^{n}) \: = \: n.x {}^{n - 1} }}$

$\implies \: v \: = \frac{d}{dt} (2.3 {t}^{3 - 1} ) + 4 \frac{d}{dt} (1.t^{1-1}) - \frac{d}{dt} (10)$

$\bold{ \boxed{ \frac{d}{dt} (constant) \: = \: 0}}$

$\implies \: v \: = \: 6 {t}^{2} + 4- 0 \\ \\ \implies \: \pink {\underline{\bold {\boxed {v \: = 6 {t}^{2} +4}}}}$

We have ,

$\huge\bold \star \: \purple {\bold{ \boxed{ a = \frac{dv}{dt} }}}$

Where ,

• a is Acceleration
• v is Velocity
• t is time

$\implies \: a = \frac{d}{dt} (6t {}^{2}+4) \\ \\ \implies a = \frac{d}{dt} (6.2 {t}^{2 - 1})+\frac{d}{dt}(4) \\ \\ \implies \pink{ \underline{\bold {\boxed { a \: = 12t}}}}$

At t = 4 sec ,

$v = 6(4) {}^{2} +4= 6(16)+4 = 96 +4 = 100\: ms { }^{ - 1} \\ \\ a = 12(4) = 48 \: ms {}^{ - 2}$

∴ The velocity and acceleration of the body at t = 4 sec , are 100 m/s and 48 m/s²

Answer 2

Answer:

\huge\bold{\underline{\underline{Given:-}}}

Given:−

Displacement of a body along a straight line is described by the equation is x = 2t³ + 4t - 10

(Here i have mentioned the term as displacement because the body is moving along strainght line which is nothing but the shortest path of the motion)

\huge\bold{\underline{\underline{To\:Find:-}}}

ToFind:−

Velocity and acceleration of the body at t = 4 sec

\huge\bold{\underline{\underline{Solution:-}}}

Solution:−

We know that ,

\huge\bold \star \: \purple { \bold{ \boxed{ v \: = \: \frac{dx}{dt} }}}⋆

v=

dt

dx

Where ,

V is velocity

x is displacement

t is time

\begin{gathered} \implies \: v \: = \: \frac{d}{dt} (2t {}^{3} + 4t - 10) \\ \\ \implies \: v = \frac{d}{dt} (2t {}^{3} ) + \frac{d}{dt} (4t) - \frac{d}{dt} (10) \end{gathered}

⟹v=

dt

d

(2t

3

+4t−10)

⟹v=

dt

d

(2t

3

)+

dt

d

(4t)−

dt

d

(10)

\bold {\boxed{ \frac{d}{dx} (a {}^{n}) \: = \: n.x {}^{n - 1} }}

dx

d

(a

n

)=n.x

n−1

\implies \: v \: = \frac{d}{dt} (2.3 {t}^{3 - 1} ) + 4 \frac{d}{dt} (1.t^{1-1}) - \frac{d}{dt} (10)⟹v=

dt

d

(2.3t

3−1

)+4

dt

d

(1.t

1−1

)−

dt

d

(10)

\bold{ \boxed{ \frac{d}{dt} (constant) \: = \: 0}}

dt

d

(constant)=0

\begin{gathered} \implies \: v \: = \: 6 {t}^{2} + 4- 0 \\ \\ \implies \: \pink {\underline{\bold {\boxed {v \: = 6 {t}^{2} +4}}}} \end{gathered}

⟹v=6t

2

+4−0

⟹

v=6t

2

+4

We have ,

\huge\bold \star \: \purple {\bold{ \boxed{ a = \frac{dv}{dt} }}}⋆

a=

dt

dv

Where ,

a is Acceleration

v is Velocity

t is time

\begin{gathered} \implies \: a = \frac{d}{dt} (6t {}^{2}+4) \\ \\ \implies a = \frac{d}{dt} (6.2 {t}^{2 - 1})+\frac{d}{dt}(4) \\ \\ \implies \pink{ \underline{\bold {\boxed { a \: = 12t}}}}\end{gathered}

⟹a=

dt

d

(6t

2

+4)

⟹a=

dt

d

(6.2t

2−1

)+

dt

d

(4)

⟹

a=12t

At t = 4 sec ,

\begin{gathered}v = 6(4) {}^{2} +4= 6(16)+4 = 96 +4 = 100\: ms { }^{ - 1} \\ \\ a = 12(4) = 48 \: ms {}^{ - 2} \end{gathered}

v=6(4)

2

+4=6(16)+4=96+4=100ms

−1

a=12(4)=48ms

−2

∴ The velocity and acceleration of the body at t = 4 sec , are 100 m/s and 48 m/s²