Question

A particle has an initial velocity of ${\mathbf{3}}\widehat {\mathbf{i}} + {\mathbf{4}}\widehat {\mathbf{j}}$ and an acceleration ${\mathbf{0}}.{\mathbf{4}}\widehat {\mathbf{i}} + {\mathbf{0}}.{\mathbf{3}}\widehat {\mathbf{j}}$ . Its speed after ${\text{10 s}}$ is:
(A) ${\text{10 units}}$
(B) ${\text{7}}\sqrt 2 {\text{ units}}$
(C) $7{\text{ units}}$
(D) ${\text{8}}{\text{.5 units}}$

Answer 1

Hint : The three equations of motion for an object with constant acceleration give us relationships between the initial velocity $\overrightarrow {\mathbf{u}}$ , velocity $\overrightarrow {\mathbf{v}}$ at a time ${\mathbf{t}}$ , constant acceleration $\overrightarrow {\mathbf{a}}$ and the distance traveled by the body ${\mathbf{s}}$ .
We have Newton’s first equation given by,
$\overrightarrow {\mathbf{v}} = \overrightarrow {\mathbf{u}} + \overrightarrow {\mathbf{a}} {\mathbf{t}}$
where, $\overrightarrow {\mathbf{u}}$ is the initial velocity of the body, $\overrightarrow {\mathbf{a}}$ is the constant acceleration of the body and $\overrightarrow {\mathbf{v}}$ is the velocity of the particle at a particular time ${\mathbf{t}}$ .

Complete Step By Step Answer:
Here, the initial velocity $\overrightarrow {\mathbf{u}} = {\mathbf{3}}\widehat {\mathbf{i}} + {\mathbf{4}}\widehat {\mathbf{j}}$ and acceleration $\overrightarrow {\mathbf{a}} = {\mathbf{0}}.{\mathbf{4}}\widehat {\mathbf{i}} + {\mathbf{0}}.{\mathbf{3}}\widehat {\mathbf{j}}$ . We need to find the velocity $\overrightarrow {\mathbf{v}}$ at time ${\mathbf{t}} = 10s$ .
Using the first equation of motion we have,
$\overrightarrow {\mathbf{v}} = ({\mathbf{3}}\widehat {\mathbf{i}} + {\mathbf{4}}\widehat {\mathbf{j}}) + ({\mathbf{0}}.{\mathbf{4}}\widehat {\mathbf{i}} + {\mathbf{0}}.{\mathbf{3}}\widehat {\mathbf{j}}) \times 10$
$\Rightarrow \overrightarrow {\mathbf{v}} = {\mathbf{3}}\widehat {\mathbf{i}} + {\mathbf{4}}\widehat {\mathbf{j}} + {\mathbf{4}}\widehat {\mathbf{i}} + {\mathbf{3}}\widehat {\mathbf{j}}$
$\Rightarrow \overrightarrow {\mathbf{v}} = {\mathbf{7}}\widehat {\mathbf{i}} + {\mathbf{7}}\widehat {\mathbf{j}}$
Now, the speed of the body is equal to the magnitude of the velocity $\overrightarrow {\mathbf{v}}$ .
So, the speed after $10s$ is ,
$v = \sqrt {{7^2} + {7^2}}$
$\Rightarrow v = \sqrt {49 + 49}$
$\Rightarrow v = \sqrt {98} = 7\sqrt 2$
So, the speed of the particle after $10s$ is $7\sqrt 2$ .
Therefore, the answer is option B. ${\text{7}}\sqrt 2 {\text{ units}}$

Additional Information:
Initial velocity is the velocity at which the particle starts to move, that is the velocity at ${\mathbf{t}} = 0{\text{s}}$ . The speed of the particle is the magnitude of the velocity $\overrightarrow {\mathbf{v}}$ , which is given as $\sqrt {{a^2} + {b^2} + {c^2}}$ for a vector $\overrightarrow {\mathbf{v}} = a\widehat {\mathbf{i}} + b\widehat {\mathbf{j}} + c\widehat {\mathbf{k}}$ .
The three equations of motion are, $(i){\text{ }}\overrightarrow {\mathbf{v}} = \overrightarrow {\mathbf{u}} + \overrightarrow {\mathbf{a}} {\mathbf{t}}$ , $(ii){\text{ }}\overrightarrow {\mathbf{s}} = \overrightarrow {\mathbf{u}} {\mathbf{t}} + \dfrac{1}{2}\overrightarrow {\mathbf{a}} {{\mathbf{t}}^2}$ and $(iii){\text{ }}2\overrightarrow {\mathbf{a}} .\overrightarrow {\mathbf{s}} = \overrightarrow {\mathbf{v}} .\overrightarrow {\mathbf{v}} - \overrightarrow {\mathbf{u}} .\overrightarrow {\mathbf{u}}$ . They are used to characterize a physical system's action in terms of its motion as a function of time.

Note :
Using the first equation here, will be less time-consuming, than using the other two equations. When the body accelerates, use the $+ ve$ sign, and when the body decelerates, use the $- ve$ sign for acceleration.