Question

The initial velocity of a particle, $\overrightarrow u = 3\overrightarrow i + 4\overrightarrow j$ . It is moving with uniform acceleration, $\overrightarrow a = 0.4\overrightarrow i + 0.3\overrightarrow j$.Its velocity after 10 seconds is?

Answer 1

Hint
We will use the first equation of motion to determine the final velocity. It is given by: -
$\overrightarrow v = \overrightarrow u + \overrightarrow a \,\Delta t$
where, $\overrightarrow v =$ final velocity of the body
$\overrightarrow u =$ initiate velocity of the body
$\overrightarrow a =$ acceleration of the body
$\Delta t =$ change in time.

Complete step by step answer
We will apply the first equation of motion to calculate the final velocity. It is expressed as below: -
$\overrightarrow v = \overrightarrow u + \overrightarrow {a\,} \Delta t$
where, $\overrightarrow v =$ final velocity of the body
$\overrightarrow u =$ initiate velocity of the body
$\overrightarrow a =$ acceleration of the body
$\Delta t =$ change in time
Now, according to question,
Initial velocity $(\overrightarrow u ) = 3\overrightarrow i + 4\overrightarrow j$)
Change in time $(\Delta t) = 10$ seconds
Acceleration $(\overrightarrow a ) = 0.4\overrightarrow i + 0.3\overrightarrow j$
Final velocity $(\overrightarrow v ) =$ to be calculated.
We will now plug these values into the formula:
$\overrightarrow v = (3\overrightarrow i + 4\overrightarrow j ) + (0.4\overrightarrow i + 0.3\overrightarrow j ) \times 10$
$\overrightarrow v = (3\overrightarrow i + 4\overrightarrow j ) + (4\overrightarrow i + 3\overrightarrow j )$
$\overrightarrow v = (3 + 4)\overrightarrow i + (4 + 3)\overrightarrow j$
$\Rightarrow \overrightarrow v = 7\overrightarrow i + 7\overrightarrow j$
Hence, the velocity after 10 seconds is $7{\text{ }}\overrightarrow i + 7\overrightarrow j$ .

Note
In this question we have used the vector form of the second equation of motion because the initial velocity and acceleration are also given in the vector form.
But we can express the final velocity in an easier to interpret way as well by finding out its magnitude and direction. Following formulas can be used to do this: -
For a vector $R = x\overrightarrow i + y\overrightarrow j$ ,
Magnitude $= |R| = \sqrt {{x^2} + {y^2}}$
Angle from horizontal (direction) $= \theta = {\operatorname{\tan} ^{ - 1}}\left( {\dfrac{y}{x}} \right)$
So now,
$|v| = \sqrt {{7^2} + {7^2}} \\\ |v| = 7\sqrt 2 \\\ \Rightarrow |v| = 9.89\dfrac{m}{s} \\\$
$\theta = {\operatorname{\tan} ^{ - 1}}\left( {\dfrac{7}{7}} \right) \\\ \theta = {\operatorname{\tan} ^{ - 1}}(1) \\\ \Rightarrow \theta = {45^\circ } \\\$
Thus, final velocity (v) has a magnitude of 9.89m/s and acts at an angle of 45 degrees from the horizontal.
This can be asked in subparts of a question like this.