Question

The displacement x of a particle varies with time t as x=4t2-15t+25. Find the position,velocity , acceleration of the particle at t=0.when will the velocity of the particle become 0?can we call the motion of the particle as one with uniform acceleration

Answer 1

position 25 m , velocity -15 m/s and acceleration 8 m/s square at t=0 .. the velocity become 0 at t =15/8 s.. and yes it is a constant acceleration .

Answer 2

Given, $x(t)=4t^{2}-15t+25$

$v(t)=\frac{dx}{dt}=8t-15\\a(t)=\frac{dv}{dt}=8$

$\therefore x(0)=25m\\v(0)=-15ms^{-1}\\a(0)=8ms^{-2}$

Also, the motion has uniform acceleration/

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