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.