Question

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 + 2.0j ) 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 ?

Answer 1

a) Velocity of the particle = 10 𝑗 ̂ 𝑚/𝑠
Acceleration of the particle =(8.0𝑖+2.0𝑗) 𝑚𝑠−2

Also,
But, a = $\frac{dv}{dt}$ = 8.0 i + 2.0 j
$\frac{dv}{dt}$= (8.0 i + 2.0 j) dt

Integrating both sides:
v(t) = 8.0 ti + 2.0 tj + u
Where,

u= Velocity vector of the particle at t = 0
v = Velocity vector of the particle at time t
But, v = $\frac{dv}{dt}$
dr = v.dt = (8.0 ti + 2.0 tj + u )dt

Integrating the equations with the conditions: at t = 0; r = 0 and at t = t; r = r
r = ut + $\frac{1}{2}$ (0.8) t2 i + $\frac{1}{2}$ x 2.0 t2j
= ut + 4.0 t2i + t1j
= (10.0 j)t + 4.0 t2 i + t2j
xi + y j = 4.0 t2i + ( 10t+ t2)j

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of 𝑖⃗ and 𝑗⃗, we get:
x=4t2
$t = (\frac{x}{4})^{\frac{1}{2}}$
and y= 10t + t2

when x = 16m:
t = $t = (\frac{16}{4})^{\frac{1}{2}}$ =2s

∴ y = 10 × 2 + (2)2 = 24 m

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b) Velocity of the particle is given by:

v(t) = 8.0 t i + 2.0 tj + u
at t=2 s
v(t) = 8.0 x 2i + 2.0 x 2j + 10j
= 16i +14j
∴ Speed of the particle:
$|v| = \sqrt{(16)^2 + (14)^2 }$
$= \sqrt{256 + 196} = \sqrt{452}$
= 21.26 m/s