NCERT solution Class 11 motion in a plane

NCERT Solutions

Class 11 Physics

Motion in a Plane

NCERT Questions

Q. 3.1
Q. 3.2
Q. 3.3
Q. 3.4
Q. 3.5
Q. 3.6
Q. 3.7
Q. 3.8
Q. 3.9
Q. 3.10
Q. 3.11
Q. 3.12
Q. 3.13
Q. 3.14
Q. 3.15
Q. 3.16
Q. 3.17
Q. 3.18
Q. 3.19
Q. 3.20
Q. 3.21
Q. 3.22

Q. 3.18

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

(a) At what time is the x- coordinates 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?

Given:-

Velocity of the particle v = 10.0m/s î

Acceleration of the particle a = (8.0 î + 2.0 ĵ) m s^-2.

a = d (v)/dt = (8.0 î + 2.0 ĵ) m s^-2

Also,

But, a = (dv/dt) = (8.0 î + 2.0 ĵ)

dv = (8.0 î + 2.0 ĵ)

Integrating both sides:

v(t) = (8.0 î + 2.0 ĵ + u)

Where,

Velocity vector of the particle at t = 0 = u

Velocity vector of the particle at time t = v

But, v = (dr/dt)

dr= v dt = (8.0 î + 2.0 ĵ + u)dt

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

But,

r = ut + (1/2)8.0t² î + (1/2) x 2.0t² ĵ

= ut + 4.0 t² î + t² ĵ

= (10.0 ĵ) t + 4.0 t² î + t² ĵ

(x î + y ĵ) = (4.0 t² î + (10t+ t²) ĵ

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of î and ĵ, we get:

x = 4t²

t =(x/4)^1/2

And y = 10t + t²

When x = 16 m:

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

Therefore, y = 10 × 2 + (2)² = 24 m

b) Velocity of the particle is given by:

v (t) = (8.0 î + 2.0t ĵ + u)

At t=2s

v (t) = 8.0 x2 î + 2.0 x2 ĵ + 10 ĵ

= (16 î + 14 ĵ)

Therefore, speed of the particle;

|v| = √ (16)² + (14)²

=√ (256 + 196) = √ (452)

=21.26m/s

Video solution:

Motion in a Plane Class 11 Physics NCERT Chapter 3 NCERT Solutions

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