Question

lf a body starts with a velocity \[2\widehat i - 3\widehat j + 11\widehat k\dfrac{m}{s}\] and moves with an acceleration of \[10\widehat i + 10\widehat j + 10\widehat k\dfrac{m}{{{s^2}}}\] then its velocity after $0.25s$ will be:
$(A)\dfrac{1}{2}\sqrt {811} \dfrac{m}{s}$
$(B)\sqrt {\dfrac{{811}}{2}} \dfrac{m}{s}$
$(C)\sqrt {811} \dfrac{m}{s}$
$(D)2\sqrt {811} \dfrac{m}{s}$

Answer 1

Hint: In this question, we will use Newton's equation of motion and then substitute the value of initial velocity, acceleration and the time taken in the equation to find the final velocity of the body. From the vector form of the answer, we will find the magnitude of the velocity obtained by the modulus of it.
The first equation of motion is given by,
\[u = 2\widehat i - 3\widehat j + 11\widehat k\dfrac{m}{s}\]
Where $v$ is the final velocity of the given body, $u$ is the initial velocity of the given body, $a$ is the acceleration of the given body and $t$ is the time taken for the change in the velocity of the given body.

Complete step by step answer:
In this question, we are given that,
\[u = 2\widehat i - 3\widehat j + 11\widehat k\dfrac{m}{s}\]
$a = 10\widehat i + 10\widehat j + 10\widehat k\dfrac{m}{{{s^2}}}$
$t = 0.25s$
Now, we will use the first equation of motion, to find the final velocity,
$v = 2\widehat i - 3\widehat j + 11\widehat k + (10\widehat i + 10\widehat j + 10\widehat k) \times 0.25$
$v = 2\widehat i - 3\widehat j + 11\widehat k + (2.5\widehat i + 2.5\widehat j + 2.5\widehat k)$
On adding similar terms,
$v = 4.5\widehat i - 0.5\widehat j + 13.5\widehat k$
The magnitude of this velocity vector is given by,
$v = \sqrt {{{(4.5)}^2} + {{(0.5)}^2} + {{(13.5)}^2}} $
On simplifying, we get,
$v = \sqrt {202.75} \dfrac{m}{s}$
$v = \dfrac{1}{2}\sqrt {811} \dfrac{m}{s}$
So, the final velocity of the given body after the time interval of $0.25s$ is obtained as $v = \dfrac{1}{2}\sqrt {811} \dfrac{m}{s}$
So, the correct answer is $(A)\dfrac{1}{2}\sqrt {811} \dfrac{m}{s}$.

Note: In the solution which is provided above, the modulus of the vector has been taken. This means that the square root is taken as the whole, outside the vector and the individual terms are separated and then squared inside the vector to find out the modulus of the vector. This modulus which has been calculated above provides the magnitude of the final velocity.