Home
Class 11
PHYSICS
A particle moves along x- axis. It's vel...

A particle moves along `x-` axis. It's velocity is a function of time according to relation `V=(3t^(2)-18t+24)m//s` assume at `t=0` particle is at origin.
Time interval in which particle speed continuous decreases?

A

`0-3 sec`

B

`0-2 sec`

C

`2-4 sec`

D

`2-3 sec`

Text Solution

Verified by Experts

The correct Answer is:
B

`V=3(t-2)(t-4)`
`a=6(t-3)`
common interval in which `V` and a both have opposite sign is `0` to `2 sec`
Promotional Banner

Topper's Solved these Questions

  • DAILY PRACTICE PROBLEMS

    RESONANCE|Exercise Dpp no 12 Physics|8 Videos
  • DAILY PRACTICE PROBLEMS

    RESONANCE|Exercise Dpp no. 13 physics|5 Videos
  • DAILY PRACTICE PROBLEMS

    RESONANCE|Exercise Dpp no 11 physics|6 Videos
  • CURRENT ELECTRICITY

    RESONANCE|Exercise Exercise|54 Videos
  • ELASTICITY AND VISCOCITY

    RESONANCE|Exercise Advanced Level Problems|9 Videos

Similar Questions

Explore conceptually related problems

A particle moves along x- axis. It's velocity is a function of time according to relation V=(3t^(2)-18t+24)m//s assume at t=0 particle is at origin. Distance travelled by particle in 0 to 3 second time interval is :

A particle moves along x- axis. It's velocity is a function of time according to relation V=(3t^(2)-18t+24)m//s assume at t=0 particle is at origin. Which of the following graph may be correct for the motion of particle

A particle is moving on a straight line with velocity (v) as a function of time (t) according to relation v = (5t^(2) - 3t + 2)m//s . Now give the answer of following questions : Acceleration of particle at the end of 3^(rd) second of motion is :

A particle is moving on a straight line with velocity (v) as a function of time (t) according to relation v = (5t^(2) - 3t + 2)m//s . Now give the answer of following questions : Velocity of particle at t = 3 sec. is :

A particle is moving on a straight line with velocity (v) as a function of time (t) according to relation v = (5t^(2) - 3t + 2)m//s . Now give the answer of following questions : Velocity of particle when acceleration is zero is :

A particle moves along x - axis in such a way that its x - co - ordinate varies with time according to the equation x=4-2t+t^(2) . The speed of the particle will vary with time as

A particle moving in a straight line has its velocit varying with time according to relation v = t^(2) - 6t +8 (m//s) where t is in seconds. The CORRECT statement(s) about motion of this particle is/are:-

A particle is moving with speed v=bsqrt(x) along positive x-axis. Calculate the speed of the particle at time t=tau (assume that the particle is at origin at t = 0).

A particle starts moving rectilinearly at time t = 0 such that its velocity v changes with time t according to the equation v = t^2-t , where t is in seconds and v is in m s^(-1) . The time interval for which the particle retrads (i.e., magnitude of velocity decreases)is

A particle starts moving rectillineraly at time t = 0 , such that its velocity 'v' changes with time 't' according to the equation v = t^(2) -t where t is in seconds and v is in m//s . The time interval for which the particle retards is