Show that `S=ut+1/2at^(2)` for uniform accelerated motion by any method

Draw velocity – time graph of uniformly accelerated motion in one dimension. From the velocity – time graph of uniform accelerated motion, deduce the equations of motion in distance and time.

Derive following equations for a uniformly accelerated motion : (i) v = u + at (ii) S = ut+(1)/(2) at^(2) (iii) v^(2)= u^(2) +2aS Where the symbols have their usual meanings .

A particle starts moving from rest such that it has uniform acceleration a_0 during the 1st second of motion, uniform acceleration (a_0)/(2) during the 2nd second of motion, uniform acceleration (a_0)/(4) during the 3rd second of motion, and so on, such that it has uniform acceleration (a_0)/(2^(n-1)) during the nth second of motion If the displacement of the particle during the 5th second of motion is D_5 =(Ka_0)/(32) , the value of K is

Assertion uniform circular motion is uniformly accelerated motion. Reason acceleration in uniform circular motion is always towards centre.

A particle shows uniform circular motion. Its motion is .

In case of a uniform circular motion, velocity and acceleration are

How is the distance related with time for the motion under uniform acceleration such as the motion of a freely falling body?

Asserion : The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. Reason : For uniform motion acceleration is zero.

A car starts from rest, attain a velocity of 36 km h^(-1) with an acceleration of 0.2 m s^(-2) , travels 9 km with this uniform velocity and then comes to halt with a uniform deaceleration of 0.1 m s^(-2) . The total time of travel of the car is

A particle starts from origin with uniform acceleration. Its displacement after t seconds is given in meter by the relation x=2+5t+7t^2 . Calculate the magnitude of its a. Initial velocity b. Velocity at t=4s c. Uniform acceleration d. Displacement at t=5s