Home
Class 12
PHYSICS
Each of the four particles move along an...

Each of the four particles move along an x axis. Their coordinates (in metres) as function of time (in seconds) are given by Particle 1 : x(t) = 3.5 - 2.7`t^3` Particle 2 : x(t) = 3.5 + 2.7`t^3` Particle 3 : x(t) = 3.5 + 2.7`t^2` Particle 4 : x(t) = 2.5 - 3.4t - 2.7 `t^2` which of these particles is speeding up for t > 0?

Promotional Banner

Similar Questions

Explore conceptually related problems

Each of four particles moves along an x axis. Their coordinates (in meters) as function of time (in seconds) are given by Particle 1:x(t) =3.5 -2.7t^(3) Particle :2 x(t)=3.5+2.7 t^(3) particle 3: x(t) =3.5+2.7 t^(2) particle 4: x(t)=3.5-3.4 t -2.7 t^(2) Which of these particles have constant acceleration ?

Position of particle moving along x-axis is given as x=2+5t+7t^(2) then calculate :

Position of particle moving along x-axis is given as x=2+5t+7t^(2) then calculate :

Position of particle moving along x-axis is given as x=2+5t+7t^(2) then calculate :

Position of particle moving along x-axis is given by x=2t^(3)-4t+3 Initial position of the particle is

A particle moves along x-axis according to the law x=(t^(3)-3t^(2)-9t+5)m . Then :-

The position of the particle moving along x-axis is given by x=2t-3t^(2)+t^(3) where x is in mt and t is in second.The velocity of the particle at t=2sec is

If the velocity of a particle moving along x-axis is given as v=(3t^(2)-2t) and t=0, x=0 then calculate position of the particle at t=2sec.

If the velocity of a particle moving along x-axis is given as v=(3t^(2)-2t) and t=0, x=0 then calculate position of the particle at t=2sec.