Figure represents position $(x)$ versus time $(t)$ graph for the motion of a particle. If $b$ and $c$ are both positive constants, which of the following expressions best describes the acceleration (A) of the particle ?

a = b + c t

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a = b - c t

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a = c t

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None of these

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Solution

The correct option is B $a = b - c t$
Given graph have 1 local maxima and 1local minima, also t of local minima is less then t of local maxima so the above graph is a cubic parabola with coefficient of $x^{3}$ negative. $- A x^{3} + B x^{2} + C x + D$
So double differentiating the abive equation we get
$- 6 A x + 2 B$
Assuming $6 A = c, 2 B = b$
We get $\frac{d^{2} x}{d t^{2}} = a c c e l e r a t i o n = a = b - c t$
So best posaible answer is option B.

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Figure represents position (x) versus time (t) graph for the motion of a particle. If b and c are both positive constants, which of the following expressions best describes the acceleration (A) of the particle ?

Figure represents position $(x)$ versus time $(t)$ graph for the motion of a particle. If $b$ and $c$ are both positive constants, which of the following expressions best describes the acceleration (A) of the particle ?