Question

A particle moves along the x axis as
x=(t -2) + a(t -2)^2, where u ,a are constants and time t is in seconds.

it's really very tough plz help me!!!!!

​ARMY PLEASE HELP!!

Answer 1

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

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

Relation given.

$\large{\bold{X = (t - 2) + a(t - 2)^2}}$

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

Taking symbol as normal meanings.

a = constant.

X = distance.

A = acceleration.

• Case-1

Differentiating w.r.t time.

will give velocity.

$\large{ V = \frac{dX}{dt}}$

$\large{ V = \frac{d( t - 2 + a (t - 2)^2)}{dt}}$

Using Product rule of Calculus.

$\large{ V = 1 - 0 + 2at}$

Again differentiating w.r.t time

will give acceleration.

$\large{ A = \frac{dV}{dt}}$

$\large{ A = \frac{d (1 + 2at)}{dt}}$

$\large{ \implies A = 2a \: m/s^2}$

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

Acceleration of the particle is 2a m/s².

• Case-2

Taking times t = 2 seconds in the distance formula.

$\large{\bold{X = (t - 2) + a(t - 2)^2}}$

$\large{ \implies X = (2 - 2) + a(2 - 2)^2}$

which will give.

$\huge{\boxed{\boxed{X = 0 meters}}}$

so, the particle will be at origin at t = 2 seconds.

Answer 2

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

From the Question,

"x" is defined by the relation:

$\sf{x = (t - 2) + a(t - 2){}^{2} }$

Differentiating x w.r.t to t,we get velocity of the particle:

$\sf{v = \frac{dx}{dt} } \\ \\ \implies \: \sf{v = \frac{d[(t - 2) + a(t - 2){}^{2}] }{dt} }$

Using Product Rule,

$\implies \: \sf{v = \frac{d(t - 2)}{dt} + \frac{d[a(t - 2) {}^{2}] }{dt} } \\ \\ \implies \: \underline{\huge{\boxed{ \sf{v = 1 + 2at}}}}$

Differentiating a w.r.t to t,we get acceleration of the particle:

$\sf{a = \frac{dv}{dt} } \\ \\ \implies \: \huge{\sf{A = 2a \: }}$

• Acceleration of the particle is 2a m/s²

Putting t = 2 in the distance function,we get:

$\sf{x = (2 - 2) + a(2 - 2){}^{2} } \\ \\ \: \huge{\rightarrow \: \sf{x = 0m}}$

As x = 0m,the particle is at origin