A body $A$ moves with a uniform acceleration $a$ and zero initial velocity. Another body $B$ , starts from the same point moves in the same direction with a constant velocity $v$ . The two bodies meet after a time $t$ . The value of $t$ is
(A) $2v/a$
(B) $v/a$
(C) $v/2a$
(D) $\sqrt {v/2a} $
Answer
302.4k+ views
Hint: We will use the equations of motion. Out of all the equations, we will select the appropriate equation connecting all the parameters. Finally, we will find the appropriate relation.
Formulae Used: \[s = ut + {\text{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2}\]
Where, $s$ is the displacement of the body, $u$ is the initial velocity of the body, $t$ is the time taken and $a$ is the acceleration of the body.
Step By Step Solution
For the body $A$ ,
\[u = 0\]
\[a = a\]
Thus, the formula turns out to be,
\[s = {\text{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2} \cdot \cdot \cdot \cdot (1)\]
Now,
For the body $B$ ,
\[u = v\]
\[a = 0\]
Thus, the formula turns out to be,
\[s = vt \cdot \cdot \cdot \cdot (2)\]
Then,
Equating $(1)$ and$(2)$, we get
\[\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2} = vt\]
\[ \Rightarrow t = 2v/a\]
Hence, the answer is (a).
Additional Information: The equations of motions helps us to connect all the parameters related to motion such as displacement, distance, speed, velocity time and acceleration. The usage of this equation depends on the given situation and the required parameter to evaluate.
The equations are:
$v = u + at$
Here, the final parameter to find the final velocity (/speed) when the initial velocity (speed), acceleration and time are given.
\[s = ut + {\text{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2}\]
Here, the final parameter to find is the displacement (in some cases distance), when the initial velocity (/speed), acceleration and time are known.
\[{v^2} - {u^2} = 2as\]
Here, the motive is to relate all the parameters.
Thus, every equation has its own purpose at the same time restrictions to be used. These equations come in very handy for solving any type of situation in motion.
Note: We took the displacement same for both the bodies. This is because they started from the same position.
Formulae Used: \[s = ut + {\text{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2}\]
Where, $s$ is the displacement of the body, $u$ is the initial velocity of the body, $t$ is the time taken and $a$ is the acceleration of the body.
Step By Step Solution
For the body $A$ ,
\[u = 0\]
\[a = a\]
Thus, the formula turns out to be,
\[s = {\text{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2} \cdot \cdot \cdot \cdot (1)\]
Now,
For the body $B$ ,
\[u = v\]
\[a = 0\]
Thus, the formula turns out to be,
\[s = vt \cdot \cdot \cdot \cdot (2)\]
Then,
Equating $(1)$ and$(2)$, we get
\[\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2} = vt\]
\[ \Rightarrow t = 2v/a\]
Hence, the answer is (a).
Additional Information: The equations of motions helps us to connect all the parameters related to motion such as displacement, distance, speed, velocity time and acceleration. The usage of this equation depends on the given situation and the required parameter to evaluate.
The equations are:
$v = u + at$
Here, the final parameter to find the final velocity (/speed) when the initial velocity (speed), acceleration and time are given.
\[s = ut + {\text{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2}\]
Here, the final parameter to find is the displacement (in some cases distance), when the initial velocity (/speed), acceleration and time are known.
\[{v^2} - {u^2} = 2as\]
Here, the motive is to relate all the parameters.
Thus, every equation has its own purpose at the same time restrictions to be used. These equations come in very handy for solving any type of situation in motion.
Note: We took the displacement same for both the bodies. This is because they started from the same position.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Understanding Uniform Acceleration in Physics

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2026-27 Free PDF Download (Sign-in Required)

