The correct option is A 3√5 m/s, θ=tan−1(2)
Given that , →r=3t^i+√3t2^j−4^k.
We know that,velocity is derivative of position with respect to time,
→v=d→rdt=3^i+2√3t^j
Now to find the magnitude and direction of velocity at t=√3 s we put the value in the above equation
→v(t=√3 s)=3^i+6^j,
magnitude of the vector a^i+b^j is given by √a2+b2 and direction is given by tanθ=ba.
Hence,
|→v|=√32+62=3√5 m/s
θ=tan−1(63)=tan−1(2)