Home
Class 9
PHYSICS
Obtain a relation for the distance trave...

Obtain a relation for the distance travelled by an object moving with a unifrom acceleration in the interval between 4th and 5th seconds.

Text Solution

Verified by Experts

From second equation of motion
Distance travelled in t see
`s=ut +1/2at^(2)`
Distance travelled in 4 s
`s_(4) = u xx 4 + 1/2a(4)^(2) [because put t = 4 s]`
`= 4u + 1/2 xx a xx 16 = 4u+8a` (`S_(4)`= distance trvelled in 4th sec)
Again, distance travelled in 5 s
`s_(5) = ut + 1/2at^(2) [because put t = 5s]`
`= uxx5 + 1/2 a (5)^(2) = 5u + (25)/(2) a (S_(5)` = distance travelled in 5th sec)
So, distance travelled in the interval between 4th and 5th second.
`s = s_(5)-s_(4)=(5u+(25)/(2)a)-(4u+8a)`
`= 5u + (25)/(2)a-4u-8a`
`= 5u-4u + (25)/(2)a-8a`
=u`+(25a-16a)/2=u+9/2a`
So, the ralation will be `(u+ 9//2a)`
Promotional Banner

Topper's Solved these Questions

  • MOTION

    NCERT EXEMPLAR|Exercise Short Answer Type Questions|1 Videos
  • GRAVITATION

    NCERT EXEMPLAR|Exercise Gravitation|26 Videos
  • SOUND

    NCERT EXEMPLAR|Exercise Sound|20 Videos

Similar Questions

Explore conceptually related problems

The distance travelled by a particle starting from rest and moving with an acceleration 4/3m/s2 ,in the third-second is

The distance travelled by a particle starting from rest and moving with an acceleration (4)/(3) ms^-2 , in the third second is.

A body starting from rest is moving with a uniform acceleration of 8m/s^(2) . Then the distance travelled by it in 5th second will be

Show that area under the velocity-time graph of an object moving with constant acceleration in a straight line in certain time interval is equal to the distance covered by the object in the interval.

Show that the area under the velocity-time graph of an object moving with constant acceleration in a straight line in certain time interval is equal to the distance covered by the object in that interval.

A particle starts moving with acceleration 2 m//s^(2) . Distance travelled by it in 5^(th) half second is

A body starting from rest and has uniform acceleration 8 m//s^(2) . The distance travelled by it in 5^(th) second will be

The given figure shows the position-time graph of an object. Find the ratio of the velocities at 5^(th) and 27^(th) second

A particle is moving in a straight line with initial velocity u and uniform acceleration f . If the sum of the distances travelled in t^(th) and (t + 1)^(th) seconds is 100 cm , then its velocity after t seconds, in cm//s , is.