Question

A particle moves in a straight line and its position X at time T is given by x squared is equal to 2 + t its acceleration is given by

Answer 1

Answer:-

Given:-

x² = 2 + t.

a = ?.

Explanation:-

it is given that

x² = t + 2

x = √(t + 2)

The velocity is dx/dt

$v = \frac{dx}{dt} = \frac{1}{2} {(t + 2)}^{ \frac{ - 1}{2} } \:$

Now differentiating w.r.t time again.

$a = \frac{dv}{dt} = \frac{ - 1}{4} {(t + 2)}^{\frac{ - 3}{2} } \:$

$since \: x = \sqrt{(t + 2)}$

$a = \frac{ - 1}{4} {x}^{ - 3}$

or,

$a = \frac{ - 1}{4 {x}^{3} }$

$\huge{\boxed{\boxed{a = \frac{ - 1}{4 {x}^{3}}}}}$

So, acceleration is -1/4x^3.

Answer 2

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

Here,

x is the distance travelled by the particle and t is the time taken

Now,

$\sf{x {}^{2} = 2 + t } \\ \\ \implies \: \boxed{\sf{x = (2 + t) {}^{ \frac{1}{2} } }}$

We need to find the acceleration of the particle.

Firstly,we would need velocity of the particle.

By differentiating x w.r.t to t,we get:

$\sf{v = \frac{dx}{dt} } \\ \\ \rightarrow \: \sf{v = \frac{d(2 + t){}^{ \frac{ 1}{2} } }{dt} } \\ \\ \rightarrow \: \boxed{ \sf{v =\frac{-1{(t+2}^{\frac{1}{2}}}{2}}}$

Now,

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

$\sf{a = \frac{dv}{dt}} \\ \\ \rightarrow \: \sf{a = \frac{ - 1(t + 2){}^{ \frac{ - 3}{2} } }{4} } \\ \\ \rightarrow \: \sf{a = \frac{ - 1( \sqrt{t + 2)}{}^{ - 3}}{4} } \\ \\ \rightarrow \: \underline{ \boxed{\sf{a = \frac{ - 1}{4x {}^{3}}}}}$

Acceleration of the particle is about -1/4x³ m/s²