A particle is thrown with any velocity vertically upward, the distance travelled by the particle in first second of its decent is

A particle is thrown vertically upward with speed 45 m//s . Find the distance covered by the particle in the 5^(th) second of its journey.

A particle is thrown up with an initial velocity such that it takes more than one second to reach the top point. What is the distance travelled by the particle during the first second of its decent?

The equation for the motion of a particle is v = at . The distance travelled by the particle in the first 4 second is :

A particle is projected vertically upward and moves freely under gravity.If the distance travelled by the particle in the 4^(th) second of its motion is twice of that in the 5^( th ) second,then the speed of projection is ?[g=10m/ s^(2) ]

A particle moves in a straight line from origin. Its velocity time curve is shown. Find the distance of particle at the =8sec. From the origin and the distance travelled by the particle during first 8 seconds.

A particle moves in a straight line from origin. Its velocity time curve is shown. Find the distance of particle at the =8sec. From the origin and the distance travelled by the particle during first 8 seconds.

A particle moves in a straight line from origin. Its velocity time curve is shown. Find the distance of particle at the =8sec. From the origin and the distance travelled by the particle during first 8 seconds.

The velocity of a particle varies with time as per the figure shown. The distance travelled by the particle in 10 seconds is ?(m/s)

The variation of velocity of particle moving along a straight line is shown in the figure. The distance travelled by the particle in 4th second is?