||τ||=13 Nm
The torque τ is in the direction of the unit vector +^k
The velocity of the body at t=1 s is v=12(^i+2^j) ms⁻¹
The magnitude of displacement of the body at t=1 s is 16 m
The correct options are A ||τ||=13 Nm C The velocity of the body at t=1 s is v=12(^i+2^j) ms⁻¹
F=(αt)^i+β^j [At t=0, v=0, r=0]
α=1, β=1
F=t^i+^j
mdv/dt=t^i+^j
On integrating we get
mv=t²/2^i+t^j [m=1 kg]
dr/dt=t²/2^i+t^j [r=0 at t=0]
On integrating we get
r=t³/6^i+t²/2^j
At t=1 sec, τ=(r×F)=(1/6^i+1/2^j)×(^i+^j)
τ = (1/6)k - (1/2)k = -(1/3)k
||τ|| = 1/3 Nm
At t=1s, v = 1/2^i+^j, ||v||=√(1/4+1)=√5/2 ≈ 1.118 ms⁻¹ (Option C is incorrect)
At t=1s, r = 1/6^i+1/2^j, ||r||=√(1/36+1/4)=√10/6 ≈ 0.527 m (Option D is incorrect)