Question

Example 3.22 The motion of a particle along a straight line is
described by the function x = (2t - 3)2 where x is in metre and
is in second
(1) Find the position, velocity and acceleration at t=2 s.
(it) Find velocity of the particle at origin.​

Answer 1

$\huge{\underline{\underline{\sf{Answer \colon}}}}$

a)

From the Question,

$\sf{x = (2t - 3) {}^{2} } \\ \\ \implies \: \boxed{\sf{x = 4t {}^{2} - 12t + 9 }}$

Here,the position of the particle is given by,x = 4t² - 12t + 9

When t = 2,

$\sf{x = 4(2) {}^{2} - 12(2) + 9} \\ \\ \implies \: \: \sf{x = 1m}$

For Velocity of the Particle,

•Differentiating x w.r.t t,we get:

$\sf{v = \frac{dx}{dt} } \\ \\ \implies \: \sf{v = \frac{d(4t {}^{2} - 12t + 9) }{dt} } \\ \\ \implies \: \boxed{ \sf{v = 8t - 12}}$

Now,

At t = 2,

$\sf{v = 8(2) - 12} \\ \\ \implies \: \sf{v = 4ms {}^{ - 1} }$

For Acceleration of the Particle,

•Differentiating v w.r.t to t,we get:

$\sf{a = \frac{dv}{dt} } \\ \\ \implies \: \sf{a = \frac{d(8t - 12)}{dt}} \\ \\ \implies \: \boxed{\sf{a = 8ms {}^{ - 2} }}$

b) Position of the particle at origin is (0,0)

Implies,

x = 0m

Now,

$\sf{v = \frac{dx}{dt} = 0ms {}^{ - 1} }$

Thus,the velocity of the particle at origin is zero

Differentiation

$\sf{nx {}^{n - 1} }$

• This expression is used to differentiate any two values

• Differentiation of zero or any constant is zero

Answer 2

$\huge{\mathfrak{\underline{\red{Answer :-}}}}$

Given :-

Time(t) = 2s

x = (2t - 3)²

Solution :-

Case 1 :- (a)

x = (2t - 3)²

Using Identity :-

$\huge{\boxed{\boxed{\red{(a - b)^{2} = a^{2} + b^{2} - 2ab}}}}$

______________[Put Values]

x = 2(2 * 2)² + (3)² - 2(2 * 2)(3)

x = 16 + 9 - 24

x = 25 - 24

x = 1 m

$\huge{\boxed{\orange{x = 1m}}}$

(b) Velocity

$\huge{\boxed{\boxed{\red{v = \frac{dx}{dt}}}}}$

_____________[Put Values]

v = d (4t² - 12t + 9)/ dt

v = 8t - 12 ( Put value of t)

v = 8(2) - 12

v = 16 - 12

v = 4 m/s

$\huge{\boxed{\orange{v = {4ms}^{ - 1} }}}$

(c)Acceleration

$\huge{\boxed{\boxed{\red{a = \frac{dx}{dt}}}}}$

a = d(8 - 12)/dt

a = 8m/s²

$\huge{\boxed{\orange{a = {8ms}^{ - 2} }}}$

Case 2 :-

Position of the particle at origin will be 0.

It implies that,

$\sf{ v = \frac{dx}{dt} = 0 {ms}^{ - 1} }$

So, the velocity of particle at origin. will be 0.