Question

Displacement of particle along x axis is given by x=4+9t+8tsquared. Find its velocity and acceleration at t =1 second

Answer 1

AnswEr :

• Velocity = 25 m/s

• Acceleration = 16 m/s²

Explanation :

Given,

Displacement of the particle :

$\sf x = 4 + 9t + 8t {}^{2}$

To finD

Velocity and Acceleration at t = 1s

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

$\sf \: v = \dfrac{dx}{dt} \\ \\ \leadsto \sf \: v = \dfrac{d(4 + 9t + {8t}^{2} )}{dt} \\ \\ \leadsto \boxed{ \boxed{\sf \: v =( 16t + 9)\: {ms}^{ - 1} }}$

Putting t = 1s,we get :

$\large{ \leadsto \sf \: v = 25 \: {ms}^{ - 1}}$

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

$\sf \: a = \dfrac{dv}{dt} \\ \\ \leadsto \sf \: a = \dfrac{d(16t + 9)}{dt} \\ \\ \large{\leadsto \: \boxed{ \boxed{ \sf \: a = 16 {ms}^{ - 2} }}}$

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Answer 2

Answer:

v = 25 m / sec

a = 16 m / sec²

Explanation:

Given :

Displacement function :

x = 4 + 9 t + 8 t²

We are asked to find velocity and acceleration at x = 1

We know :

v = d x / d t

Diff. w.r.t. t

d x / d t = 16 t + 9 + 0

v = 16 t + 9

Velocity at x = 1

v = 16 + 9 m / sec

v = 25 m / sec

Now a = d² x / d t²

Find second order derivative of given function :

d² x / d t² = 16 + 0

a = 16 m / sec²

Hence we get required answer.