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Instantaneous velocity and acceleration
Appears in
Concepts tested here
- Differentiation of position 2
All Questions
2012 AIPMT-PRE 1 question
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The motion of a particle along a straight line is described by equation: $x = 8 + 12t - t^3$ where x is in metre and t in second. The retardation of the particle when its velocity becomes zero, is$v = \frac{dx}{dt} = 12 - 3t^2$
v = 0: $12 - 3t^2 = 0 \Rightarrow t = 2$ s
$a = \frac{d^2x}{dt^2} = -6t$
At t = 2 s: $a = -12$ $ms^{-2}$, i.e. a retardation of 12 $ms^{-2}$
2010 AIPMT-PRE 1 question
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A particle moves a distance x in time t according to equation $x = (t + 5)^{-1}$. The acceleration of particle is proportional to$x = (t + 5)^{-1}$
Velocity $v = \frac{dx}{dt} = -(t + 5)^{-2}$
Acceleration $a = \frac{dv}{dt} = 2(t + 5)^{-3}$
From the velocity, $|v| = (t + 5)^{-2}$, so $(t + 5)^{-1} = |v|^{1/2}$
$a = 2\left[(t + 5)^{-1}\right]^3 = 2\left(|v|^{1/2}\right)^3 = 2|v|^{3/2}$
So $a \propto (\text{velocity})^{3/2}$
(In terms of distance, $a = 2x^3$, which is not among the choices.)