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# 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 Solution for Physics Textbook for Class 11 (2018 to Current)
Chapter 4: Motion in a Plane
Q: 21 | Page no. 87
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) Concept: Motion in a Plane with Constant Acceleration.
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