Question

Question 4.20 The position of a particle is given by


Where t is in seconds and the coefficients have the proper units for r to be in metres.

(a) Find the v and a of the particle?

(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?

Chapter Motion In A Plane Page 87

Answer 1

v= at
a= v/t
v=a/2
where a is acceleration
v is velocity
t is time

Answer 2

r = 3t i -2t² j + 4 k
Break the position vector in it components .
e.g x = 3t
y = -2t²
z = 4
We know, velocity is change in displacement per unit time .
So, differentiate all functions with respect to time.
dx/dt = Vx = 3 m/s
dy/dt = Vy = -4t m/s
dz/dt = Vz = 0
Hence, V = Vx i + Vyj - VzK
= 3i -4tj + 0K
= 3i - 4tj

Again, differentiate with respect to time .
d²x/dt² = 0
d²y/dt² = -4
d²z/dt² = 0
Hence, acceleration = -4j m/s


(B) at t= 2 sec velocity = 3i -4×2j
= 3i - 8j
= √(3² + 8²) = √73 = 8.54 m/s
Let ∅ is the angle makes by V with x axis .
Tan∅= Vy/Vx = -8/3
∅ = tan-¹(-8/3)

Acceleration is constant e.g 4m/s in negative y-axis.