Derive an equation for position -time (S=ut+1/2at^2) relation for an object that travelled a distance 's' in time 't' under uniform acceleration 'a'

### Asked by divyapragya277 | 11th Oct, 2020, 12:29: PM

Expert Answer:

###
###
Acceleration is defined as rate of change of velocity.
hence acceleration a = (v-u)/t ..................(1)
where u and v are initial and final velocity respectively and t is time taken to change velocity from u to v.
eqn.(1) can be rewritten as, v = u+at ....................(2)
Let us asume te motion is with constant accelerated motion and its acceleration is a
Let in time t, the velocity is changed from u to v. Average velocity is (u+v)/2
distance S travelled in time t = [ (u+v)/2 ] t .......................(3)
in Eqn.(3), let us substitute for v from eqn.(2), then distance S = [ (u+u+at)/2 ] t = u t + (1/2) a t^{2}

###
Acceleration is defined as rate of change of velocity.
hence acceleration a = (v-u)/t ..................(1)
where u and v are initial and final velocity respectively and t is time taken to change velocity from u to v.
eqn.(1) can be rewritten as, v = u+at ....................(2)
Let us asume te motion is with constant accelerated motion and its acceleration is a
Let in time t, the velocity is changed from u to v. Average velocity is (u+v)/2
distance S travelled in time t = [ (u+v)/2 ] t .......................(3)
in Eqn.(3), let us substitute for v from eqn.(2), then distance S = [ (u+u+at)/2 ] t = u t + (1/2) a t^{2}

^{2}

### Answered by Shiwani Sawant | 12th Oct, 2020, 11:26: PM

## Application Videos

## Concept Videos

- Two balls are dropped to the ground from different heights. one ball is dropped two second after the other but both hit the ground at the same time if the first ball takes 6 second to reach the ground then what is the initial difference between their heights take g is equal to 10 metre per second square
- a rocket which is moving with velocity 50m/s. eject out its secondary engine and its start to accelerate with 5m/s². for next 10 seconds while entering into space. than it moves with uniform velocity. calculate this uniform velocity and the distance covered in 27 seconds. after ejection of secondary engine.
- Brake applied to a car produce a uniform retardation of 90m/s .if the car was travelling with a velocity of 27m/s.than what distance will it cover before coming to rest?
- A car is travelling with a speed of 36km/h. The driver applies the brakes and retards the car uniformly. The car is stopped in 5 seconds. Find i) Retardation of the car ii) Distance travelled before it is stopped.
- A stone is thrown vertically upward after how much time it will be at height of 5m
- A particle which is moving in a straight line with constant acceleration describes distances of 10 m and 15 m in two successive seconds. find the acceleration
- a car has uniform acceleration of 8m/s². what distance will it cover in 20 s after starting from rest.?
- a=-0.5 Speed = 90km/h find distance
- An object starting from rest travels 20m in the first 2s and 160m in the next 4s. What will be the velocity after 7s from the start?
- The water drop falls at regular intervals from a tap 9 m above the ground. The fourth drop is leaving the tap at the instant, the first drop touches the ground. How high is the third drop at that instant?

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change