Question

The relation between time t and distance x is
t = ax2 + bx where a and b are constants. The
acceleration is
(a) -2av3
(b) 2av2
(c) -2av2
(d) 2bv3​

Answer 1

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

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

$\large{\bold{t = ax^2 + bx}}$

a, and b are constants.

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

$\large{t = ax^2 + bx}$

Differentiating w.r.t "x".

$\large{ \frac{dt}{dx} = \frac{d(ax^2 + bx)}{dx}}$

It becomes,

$\large{ \frac{dt}{dx} = 2ax + b}$

Taking reciprocal of the obtained value,

$\large{ \frac{dx}{dt} = \frac{1}{2ax + b}}$

as we know dx/dt is velocity,

$\large{\boxed{v = \frac{1}{2ax + b}}}$

Now, Acceleration,

$\large{\bold{a = \frac{dv}{dt}}}$

Now,

$\large{a = \frac{dv}{dt} \times \frac{dx}{dx}}$

Rearranging,

$\large{a = \frac{dx}{dt} \times \frac{dv}{dx}}$

$\large{\bold{a = v.\frac{dv}{dx}}}$

Substituting the values,

$\large{a = \left[ \frac{1}{2ax + b} \right]. \left[ \frac{d(2ax + b)^{-1}}{dx} \right]}$

Differentiating,

$\large{a = \frac{1}{2ax + b} \times -1 \times [2a (2ax + b)^{-2}]}$

$\large{a = \frac{1}{2ax + b} \times -1 \times [2a \frac{1}{ (2ax + b)^{2}}]}$

It becomes,

$\large{a = - 2a \times \frac{1}{(2ax + b)^3}}$

As we know,

$\large{v = \frac{1}{2ax + b}}$

Substituting,

$\huge{\boxed{\boxed{a = - 2av^3}}}$

So, the acceleration is - 2av³ m/s² (Option- 1).

Note:-

Here negative sign indicates the body is decelerating.

Answer 2

Answer:

The  acceleration is  $\bf -2av^3$

$\bigstar\;\boxed{\bf -2av^3}}$

Explanation:

Given -

The relation between time t and distance x is given by, t = ax² + bx, where a and b are constants.

To Find -

The Accelration.

Solution -

• First step -

Differentiate t with respect to x.

So,

$\sf\\ \\\implies \dfrac{dt}{dx} = \dfrac{d(ax^2 + bx)}{dx}\\ \\ \\ \implies \dfrac{dt}{dx}= 2ax + b\quad .....(1)$

$\sf\\ \\\implies v = \dfrac{dt}{dx}$

$\sf\\ \\\implies \dfrac{dt}{dx} = \dfrac{1}{{\dfrac{dx}{dt}}}\quad [From\;1]\\ \\ \\ \implies \dfrac{1}{v} = 2ax + b$

$\sf \implies v = \dfrac{1}{(2ax + b)}\quad.....(2)$

• Second Step -

Differentiate v with with respect to time t.

So,

⇒ dv/dt = d{1/(2ax + b)}/dt

⇒ - 1/(2ax + b)² × d(2ax + b)/dt

⇒ -1/(2ax + b)² × [2a × dx./dt ]

⇒ -1/(2ax + b)² × 2a v

From equation (2),

⇒ dv/dt = -v² × 2av = -2av³

Now,

⇒ A = dv/dt

⇒ dv/dt = A = -2av³.

[Here negative sign represent retardation.]

∴ Option 1 is correct answer.