Question

The relation between the displacement 'x' (in metre) and the time 't' (in second) of a particle is \( t = 2x^2+3x \). If the displacement of the particle is 25 cm from the origin (x=0), then the acceleration of the particle is

• A. \( +\frac{1}{16} \text{ms}^{-2} \)
• B. \( -\frac{1}{16} \text{ms}^{-2} \)
• C. \( +\frac{1}{8} \text{ms}^{-2} \)
• D. \( -\frac{1}{8} \text{ms}^{-2} \)

Hint:

When given \(t=f(x)\) instead of \(x=f(t)\), you can find acceleration using the formula \( a = -f''(x) / [f'(x)]^3 \), where \(f'(x) = dt/dx\). In this problem, \(f'(x)=4x+3\) and \(f''(x)=4\), so \(a = -4/(4x+3)^3\).

Answer 1

The Correct Option is B

We are given time as a function of displacement: \( t = 2x^2+3x \).
To find acceleration, we need to find \( \frac{d^2x}{dt^2} \). It's easier to find \( \frac{dt}{dx} \) first and then use the chain rule.
Velocity \( v = \frac{dx}{dt} \). We know that \( \frac{dx}{dt} = 1 / (\frac{dt}{dx}) \).
First, differentiate t with respect to x: \( \frac{dt}{dx} = 4x+3 \).
So, the velocity is \( v = \frac{1}{4x+3} \).
Now, we find acceleration \( a = \frac{dv}{dt} \). We can use the chain rule: \( a = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \).
Let's find \( \frac{dv}{dx} \).
\( v = (4x+3)^{-1} \).
\( \frac{dv}{dx} = -1(4x+3)^{-2} \cdot 4 = \frac{-4}{(4x+3)^2} \).
Now, calculate acceleration \(a\).
\( a = \frac{dv}{dx} \cdot v = \left( \frac{-4}{(4x+3)^2} \right) \cdot \left( \frac{1}{4x+3} \right) = \frac{-4}{(4x+3)^3} \).
We need to find the acceleration when the displacement is 25 cm.
Convert displacement to meters: \( x = 25 \text{ cm} = 0.25 \text{ m} = \frac{1}{4} \text{ m} \).
Substitute this value of x into the expression for acceleration.
\( a = \frac{-4}{(4(\frac{1}{4})+3)^3} = \frac{-4}{(1+3)^3} = \frac{-4}{4^3} = \frac{-4}{64} = -\frac{1}{16} \text{ ms}^{-2} \).