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Mathematics Question on Rate of Change of Quantities

The displacement of particle is given by x=a0+a1t2−a2t23x=a_{0}+\frac{a_{1} t}{2}-\frac{a_{2} t^{2}}{3} What is its acceleration?

A

2a23\frac{2{{a}_{2}}}{3}

B

−2a23-\frac{2{{a}_{2}}}{3}

C

a2{{a}_{2}}

D

Zero

Answer

−2a23-\frac{2{{a}_{2}}}{3}

Explanation

Solution

Key Idea Acceleration is the rate of change of velocity and velocity is the rate of change of displacement.
The displacement equation is given by
x=a0+a1t2−a2t23x=a_{0}+\frac{a_{1} t}{2}-\frac{a_{2} t^{2}}{3}
Velocity == rate of change of displacement
ie, v=dxdtv= \frac{d x}{d t}
=ddt(a0+a1t2−a2t23)=\frac{d}{d t}\left(a_{0}+\frac{a_{1} t}{2}-\frac{a_{2} t^{2}}{3}\right)
=0+a12−2a2t3=0+\frac{a_{1}}{2}-\frac{2 a_{2} t}{3}
=a12−2a2t3=\frac{a_{1}}{2}-\frac{2 a_{2} t}{3}
Acceleration = rate of change of velocity ie,
a=dvdta =\frac{d v}{d t}
=ddt(a12−2a23t)=\frac{d}{d t}\left(\frac{a_{1}}{2}-\frac{2 a_{2}}{3} t\right)
=0−2a23=0-\frac{2 a_{2}}{3}
=−2a23=-\frac{2 a_{2}}{3}