Question : A particle moves in xy plane according to the law x = 4sin6t and y = 4(1-cos6t). The distance traversed by the particle in 4 s is (x and y are in metres)
(i) 96 m
(ii) 48 m
(iii) 24 m
(iv) 108 m
Doubt by Girisha
Solution :
x = 4sin6t
differentiating both sides wrt x.
dx/dt=4cos6t×6
vx=24cos6t
y = 4(1-cos6t)
dy/dt=4(0-[-sin6t])×6
vy=24sin6t
v=√[vx²+vy²]
differentiating both sides wrt x.
dx/dt=4cos6t×6
vx=24cos6t
y = 4(1-cos6t)
dy/dt=4(0-[-sin6t])×6
vy=24sin6t
v=√[vx²+vy²]
v=√[(24cos6t)²+v(24sin6t)²]
v=√[(24)²(cos²6t+sin²6t)]
v=√[(24)²×1]
v=√[(24)²]
v=24 m/s
v=√[(24)²(cos²6t+sin²6t)]
v=√[(24)²×1]
v=√[(24)²]
v=24 m/s
Cleary, the speed of the particle is independent of time i.e. speed is constant. Hence, acceleration (a) = 0
Distance travelled by theb particle in 4sec
s=ut+½at²
s=24×4+½(0)(4)²
s=96 m
s=24×4+½(0)(4)²
s=96 m
So, (i) 96 m, would be the correct option.