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Solution - A Particle Starts from the Origin at t = 0 S with a Velocity of 10.0 JˆM/Sj^M/S and Moves in the X-y Plane with a Constant Acceleration of (8.0iˆ+2.0jˆ) - CBSE (Science) Class 11 - Physics

ConceptMotion in a Plane with Constant Acceleration

Question

A particle starts from the origin at = 0 s with a velocity of 10.0 `hatj "m/s"` and moves in the x-y plane with a constant acceleration of `(8.0 hati + 2.0 hatj) ms^(-2)`.

(a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?

(b) What is the speed of the particle at the time?

Solution 1

Velocity of the particle `vecv = 10.0 hatj` m/s

Acceleration of the particle = `veca = (8.0 hati + 2.0 hatj)`

Also

But `veca = (dvecv)/(dt) = 8.0 hati +2.0 hatj`

`(dvecv) = (8.0 hati + 2.0 hatj)dt`

Integrating both sides:

`vecv(t)= 8.0t hati + 2.0t hatj + vecu`

where

`vecu` = velocity vector of the particle at t= 0

`vecv` = velocity vector of the particle at time t

But `vecv = (dvecr)/(dt)`

`dvecr = vecvdt = (8.0t hati + 2.0t hatj + vecu)dt`

Integrating the equations with the conditions: at t = 0; r = 0 andat t = t; r = r

`vecr = vecut + 1/28.0t^2 hati + 1/2xx2.0t^2 hatj`

`=vecut + 4.0t^2 hati + t^2 hatj`

`=(10.0 hatj)t + 4.0t^2 hati + t^2 hatj`

`x hati + y hatj = 4.0t^2 hati + (10t + t^2)hatj`

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of `hati "and" hatj`, we get:

`x = 4t^2`

`t = (x/4)^(1/2)`

And `y = 10t + t^2`

(a) When x = 16 m

`t=(16/4)^(1/2)= 2s`

∴y = 10 × 2 + (2)2 = 24 m

(b) Velocity of the particle is given by:

`vecv(t) = 8.0t hati + 2.0t hatj + hatu`

at t  = 2s

`vecv(t) = 8.0 xx 2 hati + 2.0 xx 2 hatj + 10 hatj`

=`16 hati+ 14 hatj`

∴Speed of the particle

`|vecv| = sqrt((16)^2 + (14)^2)`

`=sqrt(256+196) = sqrt(452)`

`= 21.26 m/s`

Solution 2

it is given that `vecr_(t = 0s) = vecv_(0) = 10.0 hatj` m/s and `veca(t) = (8.0 hati + 2.0 hatj)   ms^(-2)`

(a) it means `x_0 = 0,u_x = 0, a_x = 8.0` `ms^(-2)` and x = 16 m

Using relation `s = x - x_0 = u_xt+1/2a_xt^2` we have

`16 - 0 = 0 + 1/2 xx 8.0 xx t^2 => t = 2s`

`:.y = y_0 + u_yt+ 1/2a_yt^2 = 0 + 10.0xx2+1/2xx2.0xx(2)^2`

= 20 + 4 = 24 m

(b) Velocity of particle at t= 2 s along x-axis

`v_x = u_x+a_xt=0 + 8.0 xx 2 = 16.0` m/s

and along y-axis `v_y = u_y+a_yt = 10.0 + 2.0 xx 2 = 14.0` m/s

∴Speed of particle at t = 2s

`v= sqrt(v_x^2+v_y^2) = sqrt((16.0)^2+(14.0)^2) = 21.26 ms^(-1)`

Is there an error in this question or solution?

APPEARS IN

 NCERT Physics Textbook for Class 11 Part 1 (with solutions)
Chapter 4: Motion in a Plane
Q: 21 | Page no. 87

Reference Material

Solution for question: A Particle Starts from the Origin at t = 0 S with a Velocity of 10.0 JˆM/Sj^M/S and Moves in the X-y Plane with a Constant Acceleration of (8.0iˆ+2.0jˆ) concept: Motion in a Plane with Constant Acceleration. For the courses CBSE (Science), CBSE (Arts), CBSE (Commerce)
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