The displacement of a particle moving in...

The displacement of a particle moving in straight line is given as function of time as `s = ((t^(3))/(3) - (3t^(2))/(2) + 2t), s` is in m and t is in sec. The particle comes to momentary rest n times Find the value of `n`

Text Solution

Verified by Experts

The correct Answer is:

`s = (t^(3))/(3) - (3t^(2))/(2) + 2t`
`:. V = (ds)/(dt) = t^(2) - 3 t + 2`
Since the particle comes to rest, therefore,
`t^(2) - 3 t + 2 = 0`
`(t - 1) (t - 2) = 0`
` t = 1 s " and " 2s`
Hence, the particle comes to rest twice.
`:. n = 2`

Topper's Solved these Questions

MOTION IN A STRAIGHT LINE
GRB PUBLICATION|Exercise Comprehension type Queston|14 Videos
MOTION IN A STRAIGHT LINE
GRB PUBLICATION|Exercise Matrix -match|4 Videos
FRICTION AND CIRCULAR MOTION
GRB PUBLICATION|Exercise Comprehension type|11 Videos
MOTION IN TWO AND THREE DIMENSIONS
GRB PUBLICATION|Exercise All Questions|210 Videos

The position of a particle moving on a straight line depends on time t as x=(t+3)sin (2t)

Its velocity at the initial moment is 6 m/s

Its acceleration at the initial is `-8 m//s^(2)`

It has velocity of 3 m/s at `t=(pi)/2` sec

It has an acceleration of `2 m//s^(2)` at `t=(pi)/4` sec

The displacement of a particle moving in a straight line, is given by s = 2t^2 + 2t + 4 where s is in metres and t in seconds. The acceleration of the particle is.

`2 m//s^2`

`4 m//s^2`

`6 m//s^2`

`8 m//s^2`

Displacement of a particle moving in a straight line is written as follows: x=(t^(3))/(3)-(5t^(2))/(2)+6t+7 what is the possible acceleration of particle when particle is in a state of rest?

`1m//s^(2)`

`-1m//s^(2)`

`-5m//s^(2)`

`+5m//s^(2)`