Question

The position x of a particle moving along a straight line at any instant is given by

 x= a0 +a1/3 +a2/4 t2. What is the acceleration?
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Answer 1

The position x of a particle moving along a straight line at any instant is given by

$x=a_0+\frac{a_1}{3}t+\frac{a_2}{4}t^2$

Its acceleration is $\boxed{\frac{a_2}{2}}$

Explanation:

Given

The position x of a particle moving along a straight line

$x=a_0+\frac{a_1}{3}t+\frac{a_2}{4}t^2$

We know that rate of change of position is velocity

i.e. $v=\frac{dx}{dt}$

And, rate of change of velocity is acceleration

i.e. $a=\frac{dv}{dt}$

Therefore,

The velocity function is given by

$v=\frac{dx}{dt}$

$\implies v=\frac{a_1}{3}+2\times\frac{a_2}{4}t$

$\implies v=\frac{a_1}{3}+\times\frac{a_2}{2}t$

And the acceleration is given by

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

$\implies a=\frac{a_2}{2}$

Hope this answer is helpful.

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