Question

The Motion Of a particle along a straight line is described by the function X=(2t-3)^3,where x is in metres and t is in seconds Find.
a)The position,velocity and acceleration at t=2s.
b)The velocity of the particle at origin.​

Answer 1

Answer:This may help you

Explanation:

X=4t^2-12t+9....(i)=>Differentiating X w.r.t t=>dX/dt=>d(4t^2-12t+9)/dt=>v=8t-12.....(ii)Differentiating v w.r.t t=>dV/dt=>d(8t-12)/dt=>a=8...(iii)Therefor,from equal...(I)=>X=1m(Ans)For equa....(ii)=>V=4m/s(Ans)As a is constant which is 8m/s^2(Ans)B. Given,(2t-3)^2=0 =>t=3/2sVelocity at time t=0 is:-=>V=0[From equa...(ii)(Ans)