A body starts from rest and acceleration 2 m per second square till it attains the maximum velocity then returnas to rest with 3 m per second square is total time is 10 seconds then maximum speed

Stopping distance formula

Object starts its motion with rest and constant acceleration then find the ratio of displacement in tsec and 2t sec

Answer in fraction

The initial velocity of a particle is 20 m per second and retardation 2m/s² then find distance and displacement in 10th sec

Ball is projected upward with velocity 50 m/s then find displacement in 6 second

The initial velocity of a particle is 20m/s and retardation 2m/s² then find distance in 10 sec

Object starts its Motion with rest and constant acceleration then find ratio of displacement in 6thsec to 6sec

Ball is projected upward with velocity 50 m/s then find time of flight

The initial velocity of a particle is 21m/s and retardation is 2 m per second square then distance in 11th second

If object starts his motion from rest and constant acceleration then its displacement and distance will be same

Write all equation of motion all six

The initial velocity of a particle is 21m/s and retardation is 2 m per second square then displacement in 11th second

Motion under gravity Draw graph for v=gt S=1/2gt² Acceleration versus time

Object starts its motion with velocity 10m/s after 8sec it's velocity becomes 24m/s then find total displacement in this interval

Acceleration does not represent motion

Ball is projected upward with velocity 50 m/s then find average speed in 8 second

Obviously start is Motion from u and constant acceleration then find velocity at mid point if velocity at end point is v

Object starts is Motion from rest and constant acceleration takes time T for S displacement then time taken for first half and second half displacement is

Ball is projected upward with velocity 50 m/s then find maximum height

Ball is projected upward with velocity 50 m/s then find average velocity in 8 second

A body falls from rest in the gravity of earth and distance travelled in 5th second is

velocity and acceleration are in opposite direction then U turn is possible only when v=0

Object starts its motion with rest and constant acceleration then find the ratio of displacement travelled by the body during the 4th sec and 3rd sec

Object starts its motion with rest and constant acceleration then find the ratio of distance in 1sec :2sec:3sec

If velocity and acceleration are in same direction then U turn is possible

Ball is projected upward with velocity 50 m/s then find height at 7 second

Objects start motion with constant acceleration a and velocity u then find its velocity at 1/3 displacement of complete journey final velocity is v

Object starts his motion from rest and constant acceleration then find ratio of displacement in nth sec to n sec

Object starts is Motion from rest and constant acceleration and travels with acceleration a for some time t1 then it retards with B and comes to at rest in time t2 then find total distance travelled and Vmax Write all for formulas of rest 2 rest motion

Ball is dropped from height H and it travelled 7H/16 in last sec of Journey, then find height H

Find displacement in the given question not distance

A body is moving with uniform acceleration describe 40 m in the first 5 second and 65 m in next 5 second its initial velocity will be

A body falls from rest in the gravity of earth and distance travelled in 45th second is