Question

A particle moves in the $x - y$ plane with only an $x$ -component of acceleration of $2m{s^{ - 2}}$ . The particle starts from the origin at $t = 0$ with an initial velocity having an $x$ -component of $8m{s^{ - 1}}$ and $y$ -component of $- 15m{s^{ - 1}}$ . Velocity of particle after time $t$ is:
(A) \left[ {\left( {8 + 2t} \right)\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} - 15\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} } \right]m{s^{ - 1}}
(B) Zero
(C) 2t\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} - 15\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j}
(D) Directed along $z$ -axis

Answer 1

Hint : To solve this question, we need to use the first kinematic equation of motion. We apply this equation separately to each of the two directions given and combine them to write the final answer.

Formula used: The formula used to solve this question is given by
$v = u + at$ , here $u$ is the initial velocity, $v$ is the final velocity, $a$ is the acceleration, and $t$ is the time.

Complete step by step answer
According to the question, at time $t = 0$ , the velocity of the particle in the $x$ -direction is equal to $8m{s^{ - 1}}$ and that in the $y$ -direction is equal to $- 15m{s^{ - 1}}$ . So the initial velocity in the vector form can be written as
\vec u = \left( {8\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} - 15\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} } \right)m{s^{ - 1}}
Also, the acceleration of the particle is equal to $2m{s^{ - 2}}$ , which is in the $x$ -direction. So the acceleration in the vector form is written as
\vec a = \left( {2\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} + 0\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} } \right)m{s^{ - 2}}
Now, from the first kinematic equation of motion we have
$v = u + at$ (1)
For $x$ -direction:
Substituting $u = 8m{s^{ - 1}}$ , $a = 2m{s^{ - 2}}$ , and $t = t$ in (1) we get
${v_x} = \left( {8 + 2t} \right)m{s^{ - 1}}$ (2)
For $y$ -direction:
Substituting $u = - 15m{s^{ - 1}}$ , $a = 0$ $t = t$ in (1) we get
${v_y} = - 15 + 0t$
$\Rightarrow {v_y} = - 15m{s^{ - 1}}$ (3)
From (2) and (3) the final velocity of the particle in vector form can be written as
v = \left[ {\left( {8 + 2t} \right)\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{i} - 15\overset{\lower0.5em\hbox{ \smash{\scriptscriptstyle\frown} }}{j} } \right]m{s^{ - 1}}
Hence, the correct answer is option A.

Note
Instead of separately solving in the $x$ and the $y$ direction, we could have directly substituted the vector forms of the initial velocity and the acceleration to get the final velocity in the vector form. The separate analysis in the $x$ and the $y$ directions is done just for the clarity. All the three kinematic equations of motion can be used in the vector form.