Motion in a plane NCERT Solutions Class 11 Physics - Solved Exercise Question 4.21

Question 4.21:

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

Velocity of the particle, v = 10.0 j m/s
Acceleration of the particle a = (8.0 i + 2.0 j)
Also,
But, a = dv / dt = 8.0 i + 2.0 j
dv = (8.0 i + 2.0 j) dt
Integrating both sides:
v(t) = 8.0t i + 2.0t j + u
where, u = Velocity vector of the particle at t = 0
v = Velocity vector of the particle at time t
But, v = dr / dt
dr = vdt = (8.0t i + 2.0t j + u) dt
Integrating the equations with the conditions: at t = 0; r = 0 and at t = t; r = r
r = ut + (1/2) X 8.0 t2 i + (1/2) X 2.0 t2 j
= ut + 4.0 t2 i + t2 j
= (10.0 j)t + 4.0t2 i +  t2 j
x i + y j = 4.0t2 i + (10t + t2) j
Since the motion of the particle is confined to the x-y plane, on equating the coefficients of i and j, we get:
x = 4t2
t = (x/4)1/2
And y = 10t + t2

(a) When x = 16 m:
t = (16/4)1/2 = 2 s
y = 10 × 2 + (2)2 = 24 m

(b) Velocity of the particle is given by:
v(t) = 8.0t i + 2.0t j + u
v(t) = 8.0t i + 2.0t j + 10 j
v(t) = 8.0t i + (10 + 2.0t) j
At t = 2 s
v(2) = 8 X 2 i + (10 + 2 X 2) j
v(2) = 16 i + 14 j
| v(2) | = ( (16)2 + (14)2 )1/2 = 21.26 m/s
Speed at t = 2 s = 21.26 m/s



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