Following are three equations of motion `S(g)=ut+(1)/(2)at^(2) v(s)=sqrt(u^(2)+2as) v(t)=u+at` Where `,S,u,t,a,v` are respectively the displacement `(` dependent variable `)`, initial `(` constant `)`, time taken `(` independent variable `)`, acceleration `(` constant `)` and final velocity `(` dependent variable `)` of the particel after time `t`. Find the displacement of the particle when its velocity becomes `10m//s` if acceleration is `5m//s^(2)` all through -
Following are three equations of motion S(g)=ut+(1)/(2)at^(2) v(s)=sqrt(u^(2)+2as) v(t)=u+at Where ,S,u,t,a,v are respectively the displacement ( dependent variable ) , initial ( constant ) , time taken ( independent variable ) , acceleration ( constant ) and final velocity ( dependent variable ) of the particel after time t . Find the velocity of the particle after 10 seconds if its acceleration is zero in interval (0 to 10s)-
Following are three equations of motion S(g)=ut+(1)/(2)at^(2) v(s)=sqrt(u^(2)+2as) v(t)=u+at Where ,S,u,t,a,v are respectively the displacement ( dependent variable ) , initial ( constant ) , time taken ( independent variable ) , acceleration ( constant ) and final velocity ( dependent variable ) of the particel after time t . Find the displacement of a particle after 10 seconds starting from rest with a uniform acceleration of 2m//s^(2)
A particle having initial velocity u moves with a constant acceleration a for a time t. a. Find the displacement of the particle in the last 1 second . b. Evaluate it for u=5m//s, a=2m//s^2 and t=10s .
A particle is moving in a straight line and its velocity varies with its displacement as v=sqrt(4+4s) m/s. Assume s=0 at t=0. find the displacement of the particle in metres at t=1s.
If displacement of particle is s=(t^(3))/(3)-(t^(2))/(2)-(t)/(2)+6 , then acceleration of the particle when its velocity is (3)/(2) , is
If velocity of a particle is v=bs^(2) ,where b is constant and s is displacement. The acceleration of the particle as function of s is?
The displacement s of a particle at a time t is given bys =t^(3)-4t^(2)-5t. Find its velocity and acceleration at t=2 .
For a particle moving along a straight line, its velocity 'v' and displacement 's' are related as v^(2) = cs , here c is a constant. If the displacement of the particle at t = 0 is zero, its velocity after 2 sec is:
RESONANCE-DAILY PRACTICE PROBLEMS-dpp 92 illustration
