1. -2/(x+2)³ 2. -2/(2x+1)³ 3. +2/(x+1)³ 4. +2/2x+1

Answer: 2. -2/(2x+1)³ Key idea: differentiate t(x) to get velocity, then use a=v dv/dx . dt/dx=2x+1⇒ v=dx/dt=1/2x+1 a=vdv/dx=1/2x+1·(-2/(2x+1)²)=-2/(2x+1)³

Topic: Frame of reference, motion in a straight line, speed and velocity· Kinematics

In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x^2 + x. The acceleration of the particle is

1. $-\frac{2}{(x+2)^3}$

2. $-\frac{2}{(2x+1)^3}$

3. $+\frac{2}{(x+1)^3}$

4. $+\frac{2}{2x+1}$

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Answer: 2. $-\frac{2}{(2x+1)^3}$

Key idea: differentiate $t(x)$ to get velocity, then use $a=v\,\dfrac{dv}{dx}$.

$$\frac{dt}{dx}=2x+1\Rightarrow v=\frac{dx}{dt}=\frac{1}{2x+1}$$$$a=v\frac{dv}{dx}=\frac{1}{2x+1}\cdot\left(-\frac{2}{(2x+1)^2}\right)=-\frac{2}{(2x+1)^3}$$

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In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x^2 + x. The acceleration of the particle is

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