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Question: Find the modulus of the mean velocity vector averaged over the first \(t\,\sec \) of motion. (A) \...

Find the modulus of the mean velocity vector averaged over the first t sec⁡t\,\sec of motion.
(A) 2a2+b2t2\sqrt {2{a^2} + {b^2}{t^2}}
(B) a2+b2t2\sqrt {{a^2} + {b^2}{t^2}}
(C) a2+bt2\sqrt {{a^2} + b{t^2}}
(D) 2a2+b2t2\sqrt {2{a^2} + {b^2}{t^2}}

Explanation

Solution

Hint The modulus of the mean velocity vector can be determined by using the formula of the mean velocity vector and the modulus of the final result is taken, then the modulus of the mean velocity vector can be determined. The formula of the mean velocity vector gives the relation between the velocity vector and the time.

Useful formula
The mean velocity vector is given by,
m⃗=∫V⃗.dt∫dt\vec m = \dfrac{{\int {\vec V.dt} }}{{\int {dt} }}
Where, m⃗\vec m is the mean velocity vector, V⃗\vec V is the velocity vector and tt is the time taken.

Complete step by step solution
The unit vector of the r⃗\vec r is given by,
r⃗=ati^−bt2j^\vec r = at\hat i - b{t^2}\hat j
The velocity vector is given by,
V⃗=dr⃗dt\vec V = \dfrac{{d\vec r}}{{dt}}
By differentiating the r⃗\vec r with respect to the time, then the velocity vector is written as,
V⃗=ai^−2btj^\vec V = a\hat i - 2bt\hat j
Now,
The mean velocity vector is given by,
m⃗=∫V⃗.dt∫dt .................(1)\vec m = \dfrac{{\int {\vec V.dt} }}{{\int {dt} }}\,.................\left( 1 \right)
By substituting the velocity vector in the above equation (1), then the above equation (1) is written as,
m⃗=∫(ai^−2btj^)dt∫dt\vec m = \dfrac{{\int {\left( {a\hat i - 2bt\hat j} \right)dt} }}{{\int {dt} }}
By integrating the above equation, then the above equation is written as,
m⃗=ati^−bt2j^t\vec m = \dfrac{{at\hat i - b{t^2}\hat j}}{t}
By cancelling the terms in the above equation, then the above equation is written as,
m⃗=ai^−btj^\vec m = a\hat i - bt\hat j
By taking modulus on the both sides, then the above equation is written as,
∣m⃗∣=∣ai^−btj^∣\left| {\vec m} \right| = \left| {a\hat i - bt\hat j} \right|
The modulus is the square root of the sum of the individual squares of the coefficient of the i^\hat i and j^\hat j, then the above equation is written as,
∣m⃗∣=a2+b2t2\left| {\vec m} \right| = \sqrt {{a^2} + {b^2}{t^2}}
Thus, the above equation shows the modulus of the mean velocity vector.

Hence, the option (B) is the correct answer.

Note The mean velocity vector is directly proportional to the integration of the velocity vector and the mean velocity vector is inversely proportional to the time. As the velocity vector increases, then the mean velocity vector also increases.