Question

A particle has initial velocity $2i + 3j$ and the acceleration is $0.3i + 0.2j$ . The magnitude of the velocity after 10 seconds will be
A. $9\sqrt 2$ units
B. $5\sqrt 2$ units
C. $5$ units
D. $9$ units

Answer 1

We can observe that the initial velocity and the acceleration are given in vector notations. So, we will break these into the x and y components and solve for the final velocity in both x and y directions. The x direction is denoted by the vector $i$ and the y direction by the vector $j$ . Then we will use the formula $v = \sqrt {{v_x}^2 + {v_y}^2}$ where ${v_x}\,,\,{v_y}$ are respective velocities in x and y directions and solve further to get the magnitude of the resultant velocity.

Formula used:
The equations of motion are
1. $v = u + at$
2. $s = ut + \dfrac{1}{2}a{t^2}$
3. $2as = {v^2} - {u^2}$
Where $u$ is the initial velocity, $v$ is the final velocity, $s$ is the distance covered, $t$ is the time taken and $a$ is the acceleration.

Complete answer:
In x direction,
The initial velocity is ${u_x} = 2\,$ units.
The acceleration is ${a_x} = 0.3$ units.
The time taken is $t = 10\,s$
Using the first speed equation, $v = u + at$ where u is the initial velocity, v is the final velocity, t is the time taken and a is the acceleration, we have
${v_x} = {u_x} + {a_x}t$
Substituting known values in the equation we get,
${v_x} = 2 + 0.3 \times 10$
Further solving the equation, we get,
${v_x} = 5$ units.
In y direction,
The initial velocity is ${u_y} = 3\,$ units.
The acceleration is ${a_y} = 0.2$ units.
The time taken is $t = 10\,s$
Using the first speed equation, $v = u + at$ where u is the initial velocity, v is the final velocity, t is the time taken and a is the acceleration, we have
${v_y} = {u_y} + {a_y}t$
Substituting known values in the equation we get,
${v_y} = 3 + 0.2 \times 10$
Further solving the equation, we get,
${v_y} = 5$ units.
Now we know that the resultant velocity will be given by the formula $v = \sqrt {{v_x}^2 + {v_y}^2}$ where ${v_x}\,,\,{v_y}$ are respective velocities in x and y directions.
Substituting the known values we get,
$v = \sqrt {{5^2} + {5^2}}$
Further solving this equation, we get,
$v = \sqrt {50}$
$\Rightarrow v = 5\sqrt 2$ units.
Hence, option (B) is correct.

Note:
Time is independent of the axis you choose. Hence, there is no component of time along any direction. It remains constant. Also, the choice of equations is important. We could have used the other two equations as well and got the same result. However, that would involve simultaneously solving many equations. The approach shown here is easier.
Alternative approach:
Instead of calculating values in x and y directions separately, we can calculate them at the same time using vectors.
$v = u + at$
Substituting the corresponding values,
$v=(2i + 3j) + (0.3i + 0.2j) \times 10$
On performing basic arithmetic operations,
$v= (2i + 3j) + (3i+2j)$
Adding these two components of $i$ and $j$
$v=5i+5j$
We need to find the magnitude which will be the same as calculated in the above solution.
$v = \sqrt {{5^2} + {5^2}}$
$\Rightarrow v = 5\sqrt 2$.