Question

A particle moves in x-y plane according to equation, x = 4t² +5t +16, y = 5t . x and y being in meter and t in second . Find the acceleration of the particle.​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Position of the particle along x - axis :

$\sf{x = 4 {t}^{2} + 5t + 16 }$

Position of the particle along y - axis :

$\sf{y = 5t}$

Along X - axis :

Differentiating x w.r.t to t,we get :

$\sf{v = \frac{dx}{dt} } \\ \\ \leadsto \: \sf{v = \frac{d(4 {t}^{2} + 5t + 18) }{dt} } \\ \\ \\ \huge{\leadsto \: \sf{v = (\: 8t + 5) \: {ms}^{ - 1} }}$

Differentiating v w.r.t to t,we get :

$\sf{a = \frac{dv}{dt} } \\ \\ \leadsto \ \sf{a = \frac{d(8t + 5)}{dt} } \\ \\ \huge{ \leadsto \: \sf{a = 8 \: {ms}^{ - 2} }}$

Along Y - axis :

Differentiating y w.r.t to t,we get :

$\sf{v = \frac{dy}{dt} } \\ \\ \leadsto \: \sf{v = \frac{d(5t)}{dt} } \\ \\ \huge{ \leadsto \: \sf{v = 5 \: {ms}^{ - 1} }}$

Differentiating v w.r.t to t,we get :

$\sf{a = \frac{dv}{dt} } \\ \\ \leadsto \: \sf{{a = \frac{d(5)}{dt} }} \\ \\ \huge{ \leadsto \: \sf{a = 0 \: {ms}^{ - 2} }}$

Since,acceleration of the particle along y - axis is zero,acceleration on x - axis is only considered

Thus,the acceleration of the particle is 8 m/s²

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A particle moves in x-y plane according to equation, x = 4t² +5t +16, y = 5t . x and y being in meter and t in second . Find the acceleration of the particle.

$\huge{\bold{\underline{\underline{Answer}}}}$

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

• x = 4t² + 5t + 16
• y = 5t
• t = Time period.

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

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Along X - Axis:-

$\large{\boxed{\tt x = 4t^2 +5t + 16}}$

Differentiating the Equation to Get velocity,

$\large{\tt \hookrightarrow v_x = \dfrac{dx}{dt}}$

$\large{\tt \hookrightarrow v_x = \dfrac{d(4t^2 + 5t + 16)}{dt}}$

$\large{\tt \hookrightarrow v_x = 8t + 5 + 0}$

$\large{\hookrightarrow {\underline{\tt v_x = 8t + 5 \: m/s}}}$

Differentiating again to get Acceleration,

$\large{\tt \hookrightarrow a_x = \dfrac{dv_x}{dt}}$

$\large{\tt \hookrightarrow a_x = \dfrac{d(8t + 5)}{dt}}$

$\large{\tt \hookrightarrow a_x = 8 + 0}$

$\large{\hookrightarrow {\underline{\tt a_x = 8 \: m/s^2}}}$

Therefore, Acceleration for x - component is 8 m/s².

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Along Y - Axis:-

$\large{\boxed{\tt y = 5t }}$

Differentiating the Equation to Get velocity,

$\large{\tt \hookrightarrow v_y = \dfrac{dy}{dt}}$

$\large{\tt \hookrightarrow v_y = \dfrac{d(5t )}{dt}}$

$\large{\tt \hookrightarrow v_y = 5 }$

$\large{\hookrightarrow {\underline{\tt v_y = 5 \: m/s}}}$

Differentiating again to get Acceleration,

$\large{\tt \hookrightarrow a_y = \dfrac{dv_y}{dt}}$

$\large{\tt \hookrightarrow a_y = \dfrac{d( 5)}{dt}}$

$\large{\tt \hookrightarrow a_y = 0}$

$\large{\hookrightarrow {\underline{\tt a_y = 0 \: m/s^2}}}$

Therefore, Acceleration for y - component is 0 m/s².

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Magnitude of Acceleration:-

$\large{\boxed{\tt a = \sqrt{(a_x)^2 + (a_y)^2}}}$

Substituting the values,

$\large{\tt \hookrightarrow a = \sqrt{(8)^2 + (0)^2}}$

$\large{\tt \hookrightarrow a = \sqrt{64 + 0}}$

$\large{\tt \hookrightarrow a = \sqrt{64}}$

$\huge{\boxed{\boxed{\tt a = 8 \: m/s^2}}}$

So, the Acceleration is 8 m/s².

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